Solving Equations with One Log Term
We can solve many equations that contain a logarithm by converting from
logarithmic form to exponential form.
To change forms we will use the following fact:
log_{b}x = L is equivalent to b^{L} = x
For example, log_{2}32 = 5 is equivalent to 2^{5} = 32.
Example 1
Solve: log_{ 8} x = 2
Solution
Rewrite in exponential form.
Simplify. 
log_{ 8} x = 2 8^{2} = x
64 = x 
So, log_{ 8 }64 = 2. This checks because 8^{2} = 64.
Note:
Notice that the equation
log_{b}x = L is NOT solved for x.
To solve for x, we rewrite it as
b^{L} = x.
Here is another way to check our solution
of log_{8}x = 2.
log_{8} x = 2
Is log_{8} 64 = 2 ?
Is log_{8} 8^{2} = 2 ? (Recall log_{b}b^{n} = n.)
Is 2 = 2 ? Yes
Example 2
Solve: ln x = 3.5. Round your answer to two decimal places.
Solution
Write ln x as loge x.
Rewrite in exponential form.
Simplify using a calculator. 
ln x =
log_{e }x =
e^{3.5} =
x ≈ 
3.5
3.5
x
33.12 
So, ln 33.12 ≈ 3.5.To check the solution, compute ln 33.12 on a calculator. The display
should read 3.50013733, which is approximately 3.5.
