Introduction to Simple Exponent and Logarithm Equations
Equations with exponents and logarithms come in many forms. Sometimes more than one strategy will work to solve them. We will first solve equations of the form “log = number” and “log = log.” We will solve an equation of the form “log = number” by rewriting the equation as an exponential equation.
Examples
Solve the equation for x .
Cancellation Law
The logarithms cancel for equations in the form “log = log” as long as the bases are the same. For example, the solution to the equation log 8 x = log 8 10 is x = 10. The cancelation law a log a x = x makes this work.
log 8 x = log 8 10
8 log 8 x = 8 log 8 10
x = 10 (By the cancelation law)
Examples
Solve for x .
log 4 = log( x − 1)
4 = x − 1 The logs cancel.
5 = x
Simple Exponent and Logarithm Equations Practice Problems
Practice
Solve for x .
- log 7 (2 x + 1) = 2
- log 4 ( x + 6) = 2
- log 5 x = 1
- log 2 (8 x − 1) = 4
- log 3 (4 x − 1) = log 3 2
- log 2 (3 − x ) = log 2 17
- ln 15 x = In( x + 4)

Solutions
- log 7 (2 x + 1) = 2
7 2 = 2 x + 1
24 = x
-
log 4 ( x + 6) = 2
4 2 = x + 6
10 = x
-
log 5 x = 1
10 1 = 5 x
2 = x
-

-

-
log 2 (3 − x ) = log 2 17
3 − x = 17
x = −14
-

-

Finding Approximate Solutions - Converting Exponential Equations to Logarithmic Equations
We need to use calculators to find approximate solutions for exponential equations whose base is e or 10. We will rewrite the exponential equation as a logarithmic equation, solve for x , and then use a calculator to get an approximate solution.
Examples
Solve for x. Give solutions accurate to four decimal places.
- e 2 x = 3
-

- 10 x +1 = 9
-

- 2500 = 1000 e x −4

Finding Approximate Solutions Practice Problems
Practice
Solve for x . Give your solutions accurate to four decimal places.
- 10 3 x = 7
- e 2 x +5 = 15
- 5000 = 2500 e 4 x
- 32 = 8 · 10 6 x −4
- 200 = 400 e −0.06 x
Solutions
- 10 3 x = 7

-
e 2 x +5 = 15

-
5000 = 2500 e 4 x

-
32 = 8 · 10 6 x −4 Divide both sides by 8.

-
200 = 400 e −0.06 x


Graphing Simple Exponent and Logarithm Equations
The logarithm function f ( x ) = log a x is the inverse of g ( x ) = a x. The graph of f ( x ) is the graph of g ( x ) with the x - and y-values reversed. To sketch the graph by hand, we will rewrite the logarithm function as an exponent equation and graph the exponent equation.
Examples
Sketch the graph of the logarithmic functions.
- y = log 2 x
-
Rewrite the equation in exponential form, x = 2 y , and let the exponent, y , be the numbers −3, −2, −1, 0, 1, 2, and 3.
Table 9.6
|
x
|
y
|
|

|
−3
|
|

|
−2
|
|

|
−1
|
|
1
|
0
|
|
2
|
1
|
|
4
|
2
|
|
8
|
3
|

Fig. 9.9
- y = ln x
-
Rewritten as an exponent equation, this is x = e y. Let y = −3, −2, −1, 0, 1, 2, and 3.
Table 9.7
|
x
|
y
|
|
0.05
|
−3
|
|
0.14
|
−2
|
|
0.37
|
−1
|
|
1
|
0
|
|
2.72
|
1
|
|
7.39
|
2
|
|
20.09
|
3
|

Fig. 9.10
As you can see by these graphs, the domain of the function f ( x ) = log a x is all positive real numbers, (0, ∞).
Graphing Simple Exponent and Logarithm Equations Practice Problems
Practice
Sketch the graph of the logarithmic function.
- y = log 1.5 x
- y = log 3 x
Solutions
-

Fig. 9.11
-

Fig. 9.12
Finding the Domain of a Function
As long as a is larger than 1, all graphs for f ( x ) = log a x look pretty much the same. The larger a is, the flatter the graph is to the right of x = 1. Knowing this and knowing how to graph transformations, we have a good idea of the graphs of many logarithmic functions.
- The graph of f ( x ) = log 2 ( x − 2) is the graph of y = log 2 x shifted to the right 2 units.
- The graph of f ( x ) = − 5 + log 3 x is the graph of y = log 3 x shifted down 5 units.
log x is the graph of y = log x flattened vertically by a factor of one-third.
The domain of f ( x ) = log a x is all positive numbers. This means that we cannot take the log of 0 or the log of a negative number. The reason is that a is a positive number. Raising a positive number to any power is always another positive number.
Examples
Find the domain. Give your answers in interval notation.
x 2 − x − 2 > 0
( x − 2)( x + 1) > 0
Put x = 2 and x = − 1 on the number line and test to see where ( x − 2) ( x + 1) > 0 is true.

Fig. 9.13
We want the “True” intervals, so the domain is (−∞, −1) ∪ (2, ∞).
Because x 2 + 1 is always positive, the domain is all real numbers, (−∞, ∞).
Finding the Domain of a Function Practice Problems
Practice
Find the domain. Give your answers in interval notation.
- f ( x ) = ln(10 − 2 x )
- h ( x ) = log( x 2 − 4)
- f ( x ) = log( x 2 + 4)
Solutions
- Solve 10 − 2 x > 0. The domain is x < 5, (−∞, 5).
- Solve x 2 − 4 > 0

Fig. 9.14
The domain is (−∞, −2) ∪ (2, ∞).
-
Because x 2 + 4 > 0 is always positive, the domain is all real numbers, (—∞, ∞).
Find practice problems and solutions for these concepts at: Exponents and Logarithms Practice Test.
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