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:
logbx = L is equivalent to bL = x
For example, log232 = 5 is equivalent to 25 = 32.
Example 1
Solve: log 8 x = 2
| Solution
Rewrite in exponential form.
Simplify. |
log 8 x = 2 82 = x
64 = x |
So, log 8 64 = 2. This checks because 82 = 64.
Note:
Notice that the equation
logbx = L is NOT solved for x.
To solve for x, we rewrite it as
bL = x.
Here is another way to check our solution
of log8x = 2.
log8 x = 2
Is log8 64 = 2 ?
Is log8 82 = 2 ? (Recall logbbn = 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 =
loge x =
e3.5 =
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.
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