4.3: Approximating the logartihm - Mathematics

A function is often approximated by its Taylor series

[f(x)=f(0)+left.x frac{d f}{d x} ight|_{x=0}+left.frac{x^{2}}{2} frac{d^{2} f}{d x^{2}} ight|_{x=0}+cdotslabel{4.28}]

which looks like an unintuitive sequence of symbols. Fortunately, pictures often explain the first and most important terms in a function approximation. For example, the one-term approximation (sin θ ≈ θ), which replaces the altitude of the triangle by the arc of the circle, turns the nonlinear pendulum differential equation into a tractable, linear equation (Section 3.5).

Another Taylor-series illustration of the value of pictures come from the series for the logarithm function:

[ln(1 + x) = x - frac{x^{2}}{2} + frac{x^{3}}{3} - ... .label{4.29}]

Its first term, x, will lead to the wonderful approximation ((1 + x)^{n} ≈ e^{nx}) for small (x) and arbitrary (n) (Section 5.3.4). Its second term, −(x^{2}/2), helps evaluate the accuracy of that approximation. These first two terms are the most useful terms—and they have pictorial explanations. The starting picture is the integral representation

[ln(1 + x) = in_{0}^{x} frac{dt}{1 + t }. label{4.30}]


What is the simplest approximation for the shaded area?

As a first approximation, the shaded area is roughly the circumscribed rectangle an example of lumping. The rectangle has area x:

[ ext{area} = underbrace{ ext{height}}_{1} x underbrace{ ext{width}}_{x} = x. label{4.31}]

This area reproduces the first term in the Taylor series. Because it uses a circumscribed rectangle, it slightly overestimates (ln(1 + x)).

The area can also be approximated by drawing an inscribed rectangle. Its width is again (x), but its height is not 1 but rather (1/(1+x)), which is approximately (1 − x) (Problem 4.18). Thus the inscribed rectangle has the approximate area (x(1 − x) = x − x). This area slightly underestimates (ln(1 + x)).

Problem 4.18 Picture for approximating the reciprocal function

Confirm the approximation

[frac{1}{1 + x} ≈ 1 − x ext{(for small x)} label{4.32}]

by trying (x = 0.1) or (x = 0.2). Then draw a picture to illustrate the equivalent approximation ((1 − x)(1 + x) ≈ 1).

We now have two approximations to (ln(1 + x)). The first and slightly simpler approximation came from drawing the circumscribed rectangle. The second approximation came from drawing the inscribed rectangle. Both dance around the exact value.


How can the inscribed- and circumscribed-rectangle approximations be combined to make an improved approximation?

One approximation overestimates the area, and the other underestimates the area; their average ought to improve on either approximation. The average is a trapezoid with area

[frac{x + (x - x^{2})}{2} = x - frac{x^{2}}{2}. label{4.33}]

This area reproduces the first two terms of the full Taylor series

[ln(1 + x) = x - frac{x^{2}}{2} + frac{x^{3}}{3} - ... label{4.34}]

Problem 4.19 Cubic term

Estimate the cubic term in the Taylor series by estimating the difference between the trapezoid and the true area.

For these logarithm approximations, the hardest problem is ln 2.

[ln (1+1) approxleft{egin{array}{ll}
1 & ext { (one term) }
1-frac{1}{2} & ext { (two terms) }
end{array} ight.label{4.35}]

Both approximations differ significantly from the true value (roughly 0.693). Even moderate accuracy for ln 2 requires many terms of the Taylor series, far beyond what pictures explain (Problem 4.20). The problem is that x in (ln(1 + x)) is 1, so the (x^{n}) factor in each term of the Taylor series does not shrink the high-n terms.

The same problem happens when computing π using Leibniz’s arctangent series (Section 4.2.3)

[arctan x = x - frac{x^{3}}{3} + frac{x^{5}}{5} - frac{x^{7}}{7} + .... label{4.36}]

By using x = 1, the direct approximation of π/4 requires many terms to attain even moderate accuracy. Fortunately, the trigonometric identity (arctan 1 = 4 arctan 1/5 − arctan 1/239) lowers the largest (x) to 1/5 and thereby speeds the convergence.


Is there an analogous that helps estimate (ln2)?

Because 2 is also (4/3)/(2/3), an analogous rewriting of (ln2) is

[ln2 = lnfrac{4}{3} - lnfrac{2}{3}. label{4.37}]

Each fraction has the form (1 + x) with (x = ±1/3). Because x is small, one term of the logarithm series might provide reasonable accuracy. Let’s therefore use (ln(1 + x) ≈ x) to approximate the two logarithms:

[ln2 ≈ frac{1}{3} - (-frac{1}{3}) = frac{2}{3}. label{4.38}]

This estimate is accurate to within 5%!

The rewriting trick has helped to compute (pi) (by rewriting the (arctan x) series) and to estimate (ln(1 + x)) (by rewriting x itself). This idea therefore becomes a method—a trick that I use twice (this definition is often attributed to Polya).

Multiple problems

Problem 4.20 How many terms?

The full Taylor series for the logarithm is

[ln(1 + x) = sum{1}^{infty} (-1)^{n+1} frac{x^{n}}{n}. label{4.39}]

If you set (x = 1) in this series, how many terms are required to estimate (ln2) to within 5%?

Problem 4.21 Second rewriting

Repeat the rewriting method by rewriting 4/3 and 2/3; then estimate (ln2) using only one term of the logarithm series. How accurate is the revised estimate?

Problem 4.22 Two terms of the Taylor series

After rewriting (ln2) as (ln(4/3) − ln(2/3)), use the two-term approximation that (ln(1+x) ≈ x − x^{2}/2) to estimate (ln2). Compare the approximation to the one-term estimate, namely 2/3. (Problem 4.24 investigates a pictorial explanation.)

Problem 4.23 Rational-function approximation for the logarithm

The replacement (ln 2 = ln(4/3) − ln(2/3)) has the general form

[ln(1 + x) = lnfrac{1 + y}{1 - y}, label{4.40}]

where (y = x/(2 + x)).

Use the expression for y and the one-term series (ln(1 + x) ≈ x) to express (ln(1 + x)) as a rational function of (x) (as a ratio of polynomials in x). What are the first few terms of its Taylor series?

Compare those terms to the first few terms of the (ln(1 + x)) Taylor series, and thereby explain why the rational-function approximation is more accurate than even the two-term series (ln(1 + x) ≈ x − x^{2}/2).

Problem 4.24 Pictorial interpretation of the rewriting

a. Use the integral representation of (ln(1 + x)) to explain why the shaded area is (ln2).

b. Outline the region that represents

[lnfrac{4}{3} - lnfrac{2}{3} label{4.41}]

when using the circumscribed-rectangle approximation for each logarithm.

c. Outline the same region when using the trapezoid approximation (ln(1 + x) = x − x^{2}/2). Show pictorially that this region, although a different shape, has the same area as the region that you drew in item b.

Lesson 14

To solve the equation (5 oldcdot e^ <3a>= 90) , Lin wrote the following:

Is her solution valid? Be prepared to explain what she did in each step to support your answer.

14.2: Natural Logarithm

  1. Complete the table with equivalent equations. The first row is completed for you.
    exponential formlogarithmic form
    a.(e^0= 1)(ln 1=0)
    b.(e^1= e)
    c.(e^ ext <-1>= frac<1>)
    d.(ln frac<1>= ext-2)
    e.(e^x = 10)
  2. Solve each equation by expressing the solution using (ln) notation. Then, find the approximate value of the solution using the “ln” button on a calculator.
    1. (e^m=20)
    2. (e^n=30)
    3. (e^p=7.5)

    14.3: Solving Exponential Equations

    Without using a calculator, solve each equation. It is expected that some solutions will be expressed using log notation. Be prepared to explain your reasoning.

    1. (10^x = 10,!000)
    2. (5 oldcdot 10^x = 500)
    3. (10^ <(x+3)>= 10,!000)
    4. (10^ <2x>= 10,!000)
    5. (10^x = 315)
    6. (2 oldcdot 10^x = 800)
    7. (10^ <(1.2x)>= 4,!000)
    8. (7oldcdot 10^ <(0.5x)>= 70)
    9. (2 oldcdot e^x=16)
    10. (10 oldcdot e^<3x>=250)
    1. Solve the equations (10^ = 16) and (10^ = 2) . Express your answers as logarithms.
    2. What is the relationship between these two solutions? Explain how you know.


    So far we have solved exponential equations by

    • finding whole number powers of the base (for example, the solution of (10^x = 100,!000) is 5)
    • estimation (for example, the solution of (10^x = 300) is between 2 and 3)
    • using a logarithm and approximating its value on a calculator (for example, the solution of (10^x = 300) is (log 300 approx 2.48) )

    Sometimes solving exponential equations takes additional reasoning. Here are a couple of examples.

    (displaystyle egin 5 oldcdot 10^x &= 45 5 oldcdot 10^x &= 4510^x &= 9x &=log 9 end )

    In the first example, the power of 10 is multiplied by 5, so to find the value of (x) that makes this equation true each side was divided by 5. From there, the equation was rewritten as a logarithm, giving an exact value for (x) .

    In the second example, the expressions on each side of the equation were rewritten as powers of 10: (10^<(0.2t)>=10^3) . This means that the exponent (0.2t) on one side and the 3 on the other side must be equal, and leads to a simpler expression to solve where we don't need to use a logarithm.

    How do we solve an exponential equation with base (e) , such as (e^x = 5) ? We can express the solution using the natural logarithm, the logarithm for base (e) . Natural logarithm is written as (ln) , or sometimes as (log_e) . Just like the equation (10^2 =100 ) can be rewritten, in logarithmic form, as (log_<10>100 = 2) , the equation (e^0 = 1) and be rewritten as (ln 1 = 0) . Similarly, (e^ < ext-2>= frac<1>) can be rewritten as (ln frac<1> = ext<->2) .

    All this means that we can solve (e^x = 5) by rewriting the equation as (x = ln 5) . This says that (x) is the exponent to which base (e) is raised to equal 5.

    To estimate the size of (ln 5) , remember that (e) is about 2.7. Because 5 is greater than (e^1) , this means that (ln 5) is greater than 1. (e^2) is about ((2.7)^2) or 7.3. Because 5 is less than (e^2) , this means that (ln 5) is less than 2. This suggests that (ln 5) is between 1 and 2. Using a calculator we can check that (ln 5 approx 1.61) .

    Glossary Entries

    The number (e) is an irrational number with an infinite decimal expansion that starts (2.71828182845 . . .) , which is used in finance and science as the base for an exponential function.

    The natural logarithm of (x) , written (ln(x)) , is the log to the base (e) of (x) . So it is the number (y) that makes the equation (e^y = x) true.

    The Illustrative Mathematics name and logo are not subject to the Creative Commons license and may not be used without the prior and express written consent of Illustrative Mathematics.

    This book includes public domain images or openly licensed images that are copyrighted by their respective owners. Openly licensed images remain under the terms of their respective licenses. See the image attribution section for more information.

    The last value above, the cube of e , is a valid solution, and often this will be all that I'm supposed to give for the answer. However, in this case (maybe leading up to graphing or word problems) they want me to provide a decimal approximation. So I plug the expression into my calculator, and round the result on my screen. My answer is:

    Note that this decimal form is not "better" than e 3 actually, e 3 is the exact, and therefore the more correct, answer. But whereas something like 2 3 can be simplified to a straight-forward 8 , the irrational value of e 3 can only be approximated in the calculator.

    Make sure you know how to operate your calculator for finding this type of solution before the next test.

    Solve log2(x) = 4.5 , accurate to two decimal places.

    This equation has a strictly-numerical term. So I'll be using The Relationship to convert the log equation to the corresponding exponential form. Then I'll solve the resulting equation.

    This requires a calculator for finding the approximate decimal value. After punching a few buttons and then rounding, I find that my answer is:

    Solving this sort of equation usually works in this way:

    If the equation has a strictly numerical term, you first use log rules to combine all log terms into one, with anything numerical on the other side of the "equals" sign. Then you use The Relationship to convert the log equation into its corresponding exponential equation, and then you may or may not meed to use your calculator to find an approximation of the exact form of the answer.

    If the equation has only log terms, then you use log rules to combine the log terms to get the equation into the form "log(of something) equals log(of something else)", and then you set (something) equal to (something else), and solve.

    By the way, when finding approximations with your calculator, don't round as you go along. Instead, do all the solving and simplification algebraically then, at the end, do the decimal approximation as one (possibly long) set of commands in the calculator. Round-off error can get really big really fast with logs, and you don't want to lose points because you rounded too early and thus too much.

    Solve log2(3x) = 4.5 , rounding your answer to two decimal places.

    This equation has a strictly numerical term, so I'll be using The Relationship to convert the log equation to its corresponding exponential form, followed by some algebra:

    If you try to check my solution above by plugging " 7.54 " into your calculator for " x " in the original equation, you will get a result that is close to 4.5 , but is not exactly equal. This is due to round-off error. That's not to say that you can't check your answers for log equation &mdash you most certainly can, and probably should &mdash but you'll need to keep this round-off-error difficulty in mind when checking your solutions. In other words, when you plug your decimal approximation into the original equation, you're just making sure that the result is close enough to be reasonable.

    For instance, to check the solution of the equation log2(3x) = 4.5, I'll plug 7.54 in for x , and see how close the result is to 4.5 :

    At this point, I'll need to use the change-of-base formula to convert this to something my calculator knows how to deal with. I'll use the natural log:

    No, the two values are not equal, but they're pretty darned close. Allowing for round-off error, these values confirm to me that I've gotten the right answer.

    If, on the other hand, my solution had returned a value of, say, 12.083 , then I would have known that, no, my answer was wrong.

    Expect to need to use a calculator for log-based word problems.

    An equation containing variables in the exponents is knowns as an exponential equation. In contrast, an equation that involves the logarithm of an expression containing a variable is referred to as a logarithmic equation.

    The purpose of solving a logarithmic equation is to find the value of the unknown variable.

    In this article, we will learn how to solve the general two types of logarithmic equations, namely:

    1. Equations containing logarithms on one side of the equation.
    2. Equations with logarithms on opposite sides of the equal to sign.

    How to solve equations with logarithms on one side?

    Equations with logarithms on one side take log b M = n ⇒ M = b n .

    To solve this type of equations, here are the steps:

    • Simplify the logarithmic equations by applying the appropriate laws of logarithms.
    • Rewrite the logarithmic equation in exponential form.
    • Now simplify the exponent and solve for the variable.
    • Verify your answer by substituting it back in the logarithmic equation. You should note that the acceptable answer of a logarithmic equation only produces a positive argument.

    Rewrite the equation to exponential form

    logs 2 (5x + 7) = 5 ⇒ 2 5 = 5x + 7

    Divide both sides by 5 to get

    Solve for x in log (5x -11) = 2

    Since the base of this equation is not given, we therefore assume the base of 10.

    Now change the write the logarithm in exponential form.

    Hence, x = 111/5 is the answer.

    Rewrite the equation in exponential form

    log10 (2x + 1) = 3n⇒ 2x + 1 = 10 3

    On dividing both sides by 2, we get

    Verify your answer by substituting it in the original logarithmic equation

    ⇒ log10 (2 x 499.5 + 1) = log10 (1000) = 3 since 10 3 = 1000

    Rewrite the equation in exponential form as

    But as you know, e = 2.718281828

    4x – 3 = (2.718281828) 3 = 20.085537

    Solve the logarithmic equation log 2 (x +1) – log 2 (x – 4) = 3

    First simplify the logarithms by applying the quotient rule as shown below.

    Now, rewrite the equation in exponential form

    Cross multiply the equation

    7x = 33 …… (Collecting the like terms)

    Solve for x if log 4 (x) + log 4 (x -12) = 3

    Simplify the logarithm by using the product rule as follows

    log 4 (x) + log 4 (x -12) = 3 ⇒ log 4 [(x) (x – 12)] = 3

    Convert the equation in exponential form.

    Since this is a quadratic equation, we therefore solve by factoring.

    x 2 -12x – 64 ⇒ (x + 4) (x – 16) = 0

    When x = -4 is substituted in the original equation, we get a negative answer which is imaginary. Therefore, 16 is the only acceptable solution.

    How to solve equations with logarithms on both sides of the equation?

    The equations with logarithms on both sides of the equal to sign take log M = log N, which is the same as M = N.

    The procedure of solving equations with logarithms on both sides of the equal sign.

    • If the logarithms have are a common base, simplify the problem and then rewrite it without logarithms.
    • Simplify by collecting like terms and solve for the variable in the equation.
    • Check your answer by plugging it back in the original equation. Remember that, an acceptable answer will produce a positive argument.

    Solve log 6 (2x – 4) + log 6 (4) = log 6 (40)

    First, simplify the logarithms.

    Solve the logarithmic equation: log 7 (x – 2) + log 7 (x + 3) = log 7 14

    Simplify the equation by applying the product rule.

    Distribute the FOIL to get

    when x = -5 and x = 5 are substituted in the original equation, they give a negative and positive argument respectively. Therefor, x = 5 is the only acceptable solution.

    Solve log 3 x + log 3 (x + 3) = log 3 (2x + 6)

    Given the equation log 3 (x 2 + 3x) = log 3 (2x + 6), drop the logarithms to get
    ⇒ x 2 + 3x = 2x + 6
    ⇒ x 2 + 3x – 2x – 6 = 0
    x 2 + x – 6 = 0……………… (Quadratic equation)
    Factor the quadratic equation to get

    4.3: Approximating the logartihm - Mathematics

    To understand what a logarithm is you first have to understand what a power is. Follow that link first if you don't!

    OK, you do know what a power is. So it makes sense to you to write something like

    After these preliminaries, we can now get into the meat of the matter. The equation (*) is the key to everything. The number b is the base , the number x the exponent , and the expression that equals y is a power . If we think of x as the independent variable and y as the dependent variable then (*) defines an exponential function .

    In the equation (*) we can now pretend that two of the variables are given, and solve for the third. If the base and the exponent are given we compute a power , if the the exponent and the power are given we compute a root (or radical ), and, if the power and the base are given, we compute a logarithm.

    In other words, The logarithm of a number y with respect to a base b is the exponent to which we have to raise b to obtain y.

    We can write this definition as

    and we say that x is the logarithm of y with base b if and only if b to the power x equals y .

    Let's illustrate this definition with a few examples. If you have difficulties with any of these powers go back to my page on powers.

    Special Bases

    More Information

    You should find extensive information on logarithms in any textbook on College Algebra. To check your understanding and guide your further study figure out answers to the following questions:

    • Why are logarithms important?
    • Why are exponential functions important?
    • How do you convert a logarithm with respect to one base to a logarithm with respect to another base?
    • Why does the base have to be positive?
    • Why is the power always positive?
    • What is it that makes natural logarithms natural?

    A Logarithm Calculator

    However, your browser does not support Java. If it did you would not see this message! Get a java compatible browser such as Netscape, of a sufficiently advanced version.

    to bring up a Logarithm Calculator that lets you pick two of the numbers in (*) and computes the third. It's pretty straightforward to use, but here is documentation.

    Logarithm Formula

    What is Logarithm?

    Now suppose if we are asked the same question but differently, like “what will be the exponent of 3 to get the result 81?”

    Then obviously answer will be 4. But how? The answer to this question only is the basic definition of logarithms.

    Now, we will write the above equation in the form of a logarithm, as

    Here, 3 is the base whose exponent we have to find. So we wish to find the value which when rose as the power to 3 will be equal to 81. Since this will be 4, so we will say that

    This above equation will be read as “log base 3 of 81 is 4”.

    Thus ,general definition and rule of logarithm is:

    Hence, the exponent or power to which a base must be raised to yield a given number is nothing but the logarithm.

    Logarithms with the base 10 are called common or Briggsian logarithms and it is written simply log n. It is invented in the 17th century to speed up calculations. The natural Logarithm is with base e where e ≅ 2.71828, and it is written as ln n.

    Logarithm Formula:

    Two most trivial identities of logarithms are:

    Some other very important formula are:

    Suppose a, b , m, n are variables with positive integers and p as a real number. Then we have,

    (1) (log_ m*n = log_ m + log_ n)

    (2) (log_ frac = log_ m + log_ n)

    (3) (log_ n^p = p log_ n)

    The word approximation is derived from Latin approximatus, from proximus meaning very near and the prefix ad- (ad- before p becomes ap- by assimilation) meaning to. [1] Words like approximate, approximately and approximation are used especially in technical or scientific contexts. In everyday English, words such as roughly or around are used with a similar meaning. [2] It is often found abbreviated as approx.

    The term can be applied to various properties (e.g., value, quantity, image, description) that are nearly, but not exactly correct similar, but not exactly the same (e.g., the approximate time was 10 o'clock).

    Although approximation is most often applied to numbers, it is also frequently applied to such things as mathematical functions, shapes, and physical laws.

    In science, approximation can refer to using a simpler process or model when the correct model is difficult to use. An approximate model is used to make calculations easier. Approximations might also be used if incomplete information prevents use of exact representations.

    The type of approximation used depends on the available information, the degree of accuracy required, the sensitivity of the problem to this data, and the savings (usually in time and effort) that can be achieved by approximation.

    Approximation theory is a branch of mathematics, a quantitative part of functional analysis. Diophantine approximation deals with approximations of real numbers by rational numbers. Approximation usually occurs when an exact form or an exact numerical number is unknown or difficult to obtain. However some known form may exist and may be able to represent the real form so that no significant deviation can be found. It also is used when a number is not rational, such as the number π, which often is shortened to 3.14159, or √ 2 to 1.414.

    Numerical approximations sometimes result from using a small number of significant digits. Calculations are likely to involve rounding errors leading to approximation. Log tables, slide rules and calculators produce approximate answers to all but the simplest calculations. The results of computer calculations are normally an approximation expressed in a limited number of significant digits, although they can be programmed to produce more precise results. [3] Approximation can occur when a decimal number cannot be expressed in a finite number of binary digits.

    Related to approximation of functions is the asymptotic value of a function, i.e. the value as one or more of a function's parameters becomes arbitrarily large. For example, the sum (k/2)+(k/4)+(k/8)+. (k/2^n) is asymptotically equal to k. Unfortunately no consistent notation is used throughout mathematics and some texts will use ≈ to mean approximately equal and

    to mean asymptotically equal whereas other texts use the symbols the other way around.

    As another example, in order to accelerate the convergence rate of evolutionary algorithms, fitness approximation—that leads to build model of the fitness function to choose smart search steps—is a good solution.

    Approximation arises naturally in scientific experiments. The predictions of a scientific theory can differ from actual measurements. This can be because there are factors in the real situation that are not included in the theory. For example, simple calculations may not include the effect of air resistance. Under these circumstances, the theory is an approximation to reality. Differences may also arise because of limitations in the measuring technique. In this case, the measurement is an approximation to the actual value.

    The history of science shows that earlier theories and laws can be approximations to some deeper set of laws. Under the correspondence principle, a new scientific theory should reproduce the results of older, well-established, theories in those domains where the old theories work. [4] The old theory becomes an approximation to the new theory.

    Some problems in physics are too complex to solve by direct analysis, or progress could be limited by available analytical tools. Thus, even when the exact representation is known, an approximation may yield a sufficiently accurate solution while reducing the complexity of the problem significantly. Physicists often approximate the shape of the Earth as a sphere even though more accurate representations are possible, because many physical characteristics (e.g., gravity) are much easier to calculate for a sphere than for other shapes.

    Approximation is also used to analyze the motion of several planets orbiting a star. This is extremely difficult due to the complex interactions of the planets' gravitational effects on each other. [5] An approximate solution is effected by performing iterations. In the first iteration, the planets' gravitational interactions are ignored, and the star is assumed to be fixed. If a more precise solution is desired, another iteration is then performed, using the positions and motions of the planets as identified in the first iteration, but adding a first-order gravity interaction from each planet on the others. This process may be repeated until a satisfactorily precise solution is obtained.

    The use of perturbations to correct for the errors can yield more accurate solutions. Simulations of the motions of the planets and the star also yields more accurate solutions.

    The most common versions of philosophy of science accept that empirical measurements are always approximations—they do not perfectly represent what is being measured.

    Symbols used to denote items that are approximately equal are wavy or dotted equals signs. [6]

    What is the radian measure of the central angle of an arc that has an arc length of 3 units and a radius of 4 units? 4/3 3/4 12 or 1 need this quick

    What ratio represents the measure of the central angle compared to the measure of the entire circle?

    If s = , what is the length of minor arc AB?

    The radius of circle(r), the arc length(s) and the angle subtended by the arc at the center of the circle(Ф) are related by the following equation:

    For the first question, we have the arc length (3 units) and the radius of circle (4 units) and we need to find the radian measure of central angle of the arc. Substituting the values in the above formula we get:

    For the second question, we have the central angle (π/2 radians) and the radius of circle(24 inches). We need to find the length of the arc. Substituting the values in above equation, we get:

    The third question is similar to the first one. The arc length is given (27 inches) and radius of circle is given to be (10 inches). We are to find the radian measure of central angle. Substituting the values in above equation, we get:

    Q.1. What are the four properties of a logarithm?
    Ans: The four basic properties of the logarithm are given below:
    (i) Product rule
    (ii) Division rule
    (iii) Power rule or the exponential rule
    (iv) Change of base rule

    Q.2. What are the properties of logarithms?
    Ans: The properties of the logarithm are given below:
    (i) Product rule: (log x + log y = log (x imes y) = log xy)
    (ii) Division rule: (left( > ight) = m – n)
    (iii) Power rule or the exponential rule: (>>left( <> ight) = n>>m)
    (iv) Change of base rule: (m = frac<<<< m>>m>><<<_a>b>>)
    (v) Base switch rule: ((a) = frac<1><<<_a>(b)>>)

    Q.3. What are the properties of logarithms and examples?
    Ans:We know that in logarithm ( = a) can be expressed as (a = y).
    (i) (<2^< – 3>> = frac<1> <8>Leftrightarrow left( <8>> ight) = – 3)
    (ii) (<10^< – 2>> = 0.01 Leftrightarrow >(0.01) = – 2)
    (iii) ( <2^6>= 64 Leftrightarrow 64 = 6)
    (iv) ( <3^2>= 9 Leftrightarrow 9 = 2)
    (v) ( <5^4>= 625 Leftrightarrow 625 = 4)
    (vi) ( <7^0>= 1 Leftrightarrow 1 = 0)
    (vii) (<3^< – 4>> = frac<1><<<3^4>>> = frac<1><<81>> Leftrightarrow frac<1><<81>> = – 4)

    Q.4. What is the power property of log?
    Ans: In this rule, the logarithm of a number (m) to the power (exponent) is equal to the exponent times its logarithm of the number (m).
    Example: (left( <> ight) = nm)

    Q.5. What is the use of logarithms?
    Ans: Logarithms are the most straightforward way to express large numbers. Mathematical calculations involving large numbers becomes easy if we use logarithm.

    Q.6. What are the principles of the logarithm?
    Ans:In exponential form, if we have ( = n), then in the logarithmic form we can write (x = n). This is the principle of working of logarithm.
    For example, ( <2^3>= 8)
    Thus, (3) is the logarithm of (8) to base (2), or (3 = 8)

    Q.7. How many types of logarithms are there?
    There are two types of logarithms, and they are as following:
    (i) Common Logarithm: The common logarithm is called the base ten logarithms. It is written as (>plog p). So, when the logarithm is taken involving base (10), we call it the common logarithm.
    Example: ( <10^2>= 100 Rightarrow >100 = 2)
    (ii) Natural Logarithm: The natural logarithm is known as the base (e) logarithm, where (e) is the Euler’s constant, which is approximately equal to (2.71828).
    The natural logarithm is written as (ln x) or (x).
    Example: ( <2.71828^4>= 54.6 Rightarrow = 54.6) or, (54.6 = 4) or simply (ln 56.6 = 4)

    We hope you find this detailed article on properties of logarithms helpful. If you have doubts or queries on this topic, feel free to ask us in the comment section and we will be ready to help you at the earliest.

    4.3: Approximating the logartihm - Mathematics

    Two kinds of logarithms are often used in chemistry: common (or Briggian) logarithms and natural (or Napierian) logarithms. The power to which a base of 10 must be raised to obtain a number is called the common logarithm (log) of the number. The power to which the base e (e = 2.718281828. ) must be raised to obtain a number is called the natural logarithm (ln) of the number.

    In simpler terms, my 8th grade math teacher always told me: LOGS ARE EXPONENTS!! What did she mean by that?

      Using log10 ("log to the base 10"):
      log10100 = 2 is equivalent to 10 2 = 100
      where 10 is the base, 2 is the logarithm (i.e., the exponent or power) and 100 is the number.

    The rest of this mini-presentation will concentrate on logarithms to the base 10 (or logs). One use of logs in chemistry involves pH, where pH = -log10 of the hydrogen ion concentration.

    Here are some simple examples of logs.

    NumberExponential ExpressionLogarithm
    100010 3 3
    10010 2 2
    1010 1 1
    110 0 0
    1/10 = 0.110 - 1 -1
    1/100 = 0.0110 - 2 -2
    1/1000 = 0.00110 - 3 -3

      Example 1: log 5.43 x 10 10 = 10.73479983. (way too many significant figures)

    So, let's look at the logarithm more closely and figure out how to determine the correct number of significant figures it should have.

      Example 1: log 5.43 x 10 10 = 10.735
      The number has 3 significant figures, but its log ends up with 5 significant figures, since the mantissa has 3 and the characteristic has 2.

      Example 4: What is the pH of an aqueous solution when the concentration of hydrogen ion is 5.0 x 10 - 4 M?

    FINDING ANTILOGARITHMS (also called Inverse Logarithm)

    1. enter the number,
    2. press the inverse (inv) or shift button, then
    3. press the log (or ln) button. It might also be labeled the 10 x (or e x ) button.

      Example 5: log x = 4.203 so, x = inverse log of 4.203 = 15958.79147. (too many significant figures)
      There are three significant figures in the mantissa of the log, so the number has 3 significant figures. The answer to the correct number of significant figures is 1.60 x 10 4 .

      Example 8: What is the concentration of the hydrogen ion concentration in an aqueous solution with pH = 13.22?


    Because logarithms are exponents, mathematical operations involving them follow the same rules as those for exponents.

    Watch the video: ΠΛΗ30 ΒΕΞΕΤΑΣΤΙΚΗ 2015-2016: ΘΕΜΑ (October 2021).