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3.5: Logarithmic Functions

Difficulty Level: At Grade Created by: CK-12
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Ears are amazing.

The average human ear can just as easily recognize the whisper of the person in the next chair as the roar of a jet plane on takeoff. Most people know that, but many people do not realize that the difference in power between the two is approximately 10,000,000,000 times!

If you look up the decibel ratings of these two sounds, you will find that the quietest whispers are apx 10db, and that a 747 on takeoff can hit 120db.

That's only a difference of 100db, how can that be the same as 10 BILLION times?

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- James Sousa: Solving Logarithmic Equations


Every exponential expression can be written in logarithmic form. For example, the equation x = 2y is written as follows: y = log2x. In general, the equation logbn = a is equivalent to the equation ba = n. That is, b is the base, a is the exponent, and n is the power, or the result you obtain by raising b to the power of a. Notice that the exponential form of an expression emphasizes the power, while the logarithmic form emphasizes the exponent. More simply put, a logarithm (or “log” for short) is an exponent.

Perhaps the most common example of a logarithm is the Richter scale, which measures the magnitude of an earthquake. The magnitude is actually the logarithm base 10 of the amplitude of the quake. That is, m = log10A . This means that, for example, an earthquake of magnitude 4 is 10 times as strong as an earthquake with magnitude 3. We can see why this is true of we look at the logarithmic and exponential forms of the expressions: An earthquake of magnitude 3 means 3 = log10A. The exponential form of this expression is 103 = A. Thus the amplitude of the quake is 1,000. Similarly, a quake with magnitude 4 has amplitude 104 = 10,000.

Solving Logarithmic Equations

In general, to solve an equation means to find the value(s) of the variable that makes the equation a true statement. To solve log equations, we have to think about what “log” means.

Consider the equation log2x = 5 . What is the exponential form of this equation?
The equation log2x = 5 means that 25 = x . So the solution to the equation is x = 25 = 32.

In some log equations, both sides of the equation contain a log. To solve these equations, use the following rule:

logb f(x) = logb g(x) → f(x) = g(x) .

In other words, set the bases equal and solve for the variable in the exponent by treating the exponents on both sides of the equation as simple polynomials.

Example A

Rewrite each exponential expression as a log expression.

a. 34 = 81 b. b4x = 52


a. In order to rewrite an expression, you must identify its base, its exponent, and its power. The 3 is the base, so it is placed as the subscript in the log expression. The 81 is the power, and so it is placed after the “log”. Thus we have: 34 = 81 is the same as log381 = 4 .
To read this expression, we say “the logarithm base 3 of 81 equals 4.” This is equivalent to saying “ 3 to the 4th power equals 81.”
b. The b is the base, and the expression 4x is the exponent, so we have:logb52 = 4x . We say, “log base b of 52, equals 4x.”

Example B

Evaluate the function f(x) = log2x for the values:

a.) \begin{align*}x = 2\end{align*}x=2 b.) \begin{align*}x = 1\end{align*}x=1 c.) \begin{align*}x = -2\end{align*}x=2


a.) If \begin{align*}x = 2\end{align*}x=2, we have:
f(x) = log2x
f(2) = log22
To determine the value of log22, you can ask yourself: “2 to what power equals 2?” Answering this question is often easy if you consider the exponential form: 2? = 2
The missing exponent is 1. So we have f(2) = log22 = 1
b.) If \begin{align*}x = 1\end{align*}x=1, we have:
f(x) = log2x
f(1) = log21
As we did in (a), we can consider the exponential form: 2? = 1. The missing exponent is 0. So we have f(1) = log21 = 0.
c. If \begin{align*}x = -2\end{align*}x=2, we have:
f(x) =log2x
f(-2) = log2 -2
Again, consider the exponential form: 2? = -2. There is no such exponent. Therefore f(-2) = log2 -2 does not exist.

Example C

Consider the equation log2(3x-1) = log2(5x - 7).

Because the logarithms have the same base (2), the arguments of the log (the expressions 3x - 1 and 5x - 7) must be equal. So we can solve as follows:

log2(3x-1) = log2(5x - 7)
3x - 1 = 5x - 7
+7 |+7
3x + 6 = 5x
6 = 2x
x = 3


Argument: The expression “inside” a logarithmic expression. The argument represents the “power” in the exponential relationship.

Exponential functions are functions with the input variable (the x term) in the exponent.

Logarithmic functions are the inverse of exponential functions. Recall: logbn = a is equivalent to ba = n.

log: The shorthand term for 'the logarithm of', as in: "logbn" = "the logarithm, base 'b', of 'n' ".

Evaluate: Identify a function value (y) for a given value of the independent variable (x).

Guided Practice

1) Solve for x: log2(9x) = log2(3x + 8)

2) Rewrite the logarithmic expressions in exponential form.

a. log10100=2 b. logbw = 5

3) Solve each equation for \begin{align*}x\end{align*}x:

a.) log4 x = 3 b.) log5 (x + 1) = 2 c.) 1 + 2log3 (x - 5) = 7

4) Write the given equation in logarithmic form: \begin{align*}5^2 = 25\end{align*}52=25


1) The log equation implies that the expressions 9x and 3x + 8 are equal:

\begin{align*}log_2(9x)\end{align*}log2(9x) \begin{align*}= log_2(3x + 8)\end{align*}=log2(3x+8)
\begin{align*}9x\end{align*}9x \begin{align*}= 3x + 8\end{align*}=3x+8
\begin{align*}-3x\end{align*}3x \begin{align*}-3x\end{align*}3x
\begin{align*}6x\end{align*}6x \begin{align*}= 8\end{align*}=8
\begin{align*}x\end{align*}x \begin{align*}= \frac{8} {6}\end{align*}=86
\begin{align*}x\end{align*}x \begin{align*}= \frac{4} {3}\end{align*}=43

2) a) The base is 10, and the exponent is 2, so we have: 102 = 100

b) The base is b, and the exponent is 5, so we have: b5 = w .

3) a) Writing the equation in exponential form gives us the solution: x = 43 = 64.

b) Writing the equation in exponential form gives us a new equation: 52 = x + 1.
We can solve this equation for x:
52 = x + 1
25 = x + 1
x = 24
c) First we have to isolate the log expression:
1 + 2log3(x - 5) = 7
2log3(x - 5) = 6
log3(x - 5) = 3
Now we can solve the equation by rewriting it in exponential form:
log3(x - 5) = 3
33 = x - 5
27 = x - 5
x = 32

4) The exponential function is in the form: \begin{align*} y = b^x\end{align*}y=bx

That means the corresponding logarithmic function is \begin{align*}x = log _b y\end{align*}x=logby

In this problem, the base \begin{align*}b\end{align*}b in the exponential function the same as the base \begin{align*}b\end{align*}b in the logarithm, so we substitute.

Substitute 2 for \begin{align*}x\end{align*}x, 5 for \begin{align*}b\end{align*}b and 25 for \begin{align*}y\end{align*}y in the logarithmic function.

This gives: \begin{align*}2 = log_5 25\end{align*}2=log525


  1. State the expression in English: \begin{align*}log_3 243\end{align*}log3243
  2. Write the given equation in logarithmic form: \begin{align*}(\frac{1}{3})^{-5} = 243\end{align*}(13)5=243
  3. Determine if the two equations below are equivalent and label them according to function family: \begin{align*}y = log_a x\end{align*}y=logax and \begin{align*}x = a^y\end{align*}x=ay

Evaluate the logarithms:

  1. \begin{align*}log_4 625\end{align*}log4625
  2. \begin{align*}log_6 64\end{align*}log664
  3. \begin{align*}log_3 216\end{align*}log3216

Evaluate each logarithm at the indicated value of x.

  1. \begin{align*}f(x) = log_2 x\end{align*}f(x)=log2x for \begin{align*}x = 32\end{align*}x=32
  2. \begin{align*}f(x) = log_3 x\end{align*}f(x)=log3x for \begin{align*}x = 1\end{align*}x=1
  3. \begin{align*}f(x) = log_4 x\end{align*}f(x)=log4x for \begin{align*}x = 2\end{align*}x=2
  4. \begin{align*}f(x) = log_{10} x\end{align*}f(x)=log10x for \begin{align*}x = \frac{1}{100}\end{align*}x=1100

Solve for x:

  1. \begin{align*}log_3 (2x+2) = log_3 (x-4)\end{align*}log3(2x+2)=log3(x4)
  2. \begin{align*}log_7 (\frac{x}{3}) = log_7 (x - 4)\end{align*}log7(x3)=log7(x4)
  3. \begin{align*}log_4 \sqrt[3]{8x} = log_4 8\end{align*}log48x3=log48
  4. \begin{align*}log_5 (x + 9) = log_5 3(x - 2)\end{align*}log5(x+9)=log53(x2)
  5. Evaluate \begin{align*}log_4 x\end{align*}log4x for \begin{align*}\frac{(x + 7)}{2} = 1204\end{align*}(x+7)2=1204

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Argument The argument of a logarithmic expression is the expression “inside” a logarithmic expression. The argument represents the “power” in the exponential relationship.
Base When a value is raised to a power, the value is referred to as the base, and the power is called the exponent. In the expression 32^4, 32 is the base, and 4 is the exponent.
Evaluate To evaluate an expression or equation means to perform the included operations, commonly in order to find a specific value.
Exponent Exponents are used to describe the number of times that a term is multiplied by itself.
Exponential Function An exponential function is a function whose variable is in the exponent. The general form is y=a \cdot b^{x-h}+k.
log "log" is the shorthand term for 'the logarithm of', as in "\log_b n" means "the logarithm, base b, of n."
Logarithmic Form Logarithmic form is \log_b a=x, such that b is the base.
Logarithmic function Logarithmic functions are the inverses of exponential functions. Recall that \log_b n=a is equivalent to b^a=n.
Power The "power" refers to the value of the exponent. For example, 3^4 is "three to the fourth power".

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Difficulty Level:
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Date Created:
Nov 01, 2012
Last Modified:
Jul 22, 2016
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