Name:
Date:
Algebra 2
Equation Solving with Exponents and Logs
Changing between exponential and logarithmic equations:
Since logarithms are defined as solutions to exponential equations, it’s possible to change
any exponential equation into a logarithmic equation, or vice versa, like this:
bx = c
 logb c = x
To get the full benefit of this property, it’s important to understand that the b, c, and x
places can be filled with different things: numbers, variables, or more complicated
expressions. You might find it easier to understand the rule this way:
OD = à
log à = D

O
where each “shape” (O, , ) can be replaced with a number, variable, or expression of
your choice. Some examples:
23 = 8  log2 8 = 3
4x = 128
 log4 128 = x
52x = 125  log5 125 = 2x
Changing back-and-forth between these two forms is so frequently helpful that it’s one of
the first things that you should try in many kinds of problems: write down the other form
and see whether it gets you closer to finding the solution.
You do need to remember which “shape” goes in which place in each equation. Here’s a
way to remember:
*One of the positions in each equation is called the “base,” and both sides have
the same base (the b, O, 2, 4, and 5 in the equations above).
*The other two shapes each get a chance to sit by themselves on one side of one
of the equations.
So long as you get those things right, everything else will fall in the right places. And it
doesn’t matter which symbols you use (b-c-x or O-- or something else that you make
up), only the positioning matters.
Name:
Date:
Algebra 2
1. Convert each of the following equations into the other form. (That is, convert each
exponent equation to a log equation and each log equation to an exponent equation.)
a. log5 216 = y
b. rs = 2t
c. log5 x = f(x)
d. w2 = 216
e. log4 (3x) = 2
f. 343 = 73
Hint: same equation as 73 = 343
g. 2 = log9 x
Hint: same equation as log9 x = 2
h. 3(2x + 1) = 81
i. log(2x)
1
3
=y
j. c = loga b
k. (4x)y = 3z
Hint: be careful, b isn’t the base in this problem
Name:
Date:
Algebra 2
2. Here are a few equations that can be solved without writing out any steps, because the
solutions are easy to find mentally. Just think of a number that you can put in for x to
make the equation true. That number is the solution.
a. 3x = 34
b. log5 18 = log5 x
c. 2x+5 = 28
Hint: The exponents on the two sides have to be equal.
d. log10 (2x) = log10 6
e. log3 10 = log3 ( 12 x)
Hint: The 10 and the
1
2
x need to be equal.
Name:
Date:
Algebra 2
The methods that worked in problem 2 can be summarized as follows:

If bM = bN then M = N. (“If there are equal powers with the same base, the exponents are
equal.”)

If logb M = logb N then M = N. (“If there are equal logs with the same base, their inputs are
equal.”)
You can use either of these ideas as a step in a multi-step problem. Here are two
examples, then a few problems for you to try.
Example: Solve 52x + 1 = 57
The exponents have to be equal:
Then finish solving:
2x + 1 = 7
2x
=6
x
=3
Example: Solve log10 (3x – 2) = log10 (16)
The inputs have to be equal:
Then finish solving:
3x – 2 = 16
3x
= 18
x
=6
3. Solve these equations using the method explained above. You should not need to use
a calculator.
a. 6(5 – x) = 62
b. log8 (2x + 3) = log8 (3x + 1)
c. 4(x + 8) = 64
Hint: First step: replace 64 with 43.
d. log4 64 = x + 8
e. 2(6x) =
1
16
Hint: First step: rewrite
1
16
as a power of 2.
Name:
Date:
Algebra 2
4. Solve each equation for x. The final answer to each problem should have the form
x = a number (which could be whole or fraction or decimal). Switching an equation
into the other form will be a helpful step on many of these problems.
The problems marked  should be done without using your calculator.
Otherwise, you may use your calculator where needed. Remember that logb c can be
evaluated as log(c)/log(b), but try to use that fact only for logs that you can’t figure
out in your head.
a. log3 x = 5

b. logx 8 = 3
c. log4 64 = 5x 
d. 10x = 200
e. 5(2x) = 125
f. 300(x – 1) = 100

Hint: Keep the (2x) together until the very last step.
Hint: Keep the (x – 1) together until the very last step.
Name:
Date:
Algebra 2
g. log9 27 = 2x + 1
h. log 3 (5x) = -4
i.
log 1 x = -2


5
j. 4000 = 3000 · (1.05)x
k.
(12 )x = 64

Hint: Must take a step involving the 3000 to get the
equation in a form that can be changed into log form.