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When solving algebraic equations, inverse operations are used to isolate
variables.
If 2 ๐ฅ + 1 = 3 , we would
subtract 1 and divide by 2 to
get the variable ๐ฅ by itself.
If
โ๐ฅโ 2
5
= 8 , we would multiply by 5 ,
then square both sides of the equation,
and finally add 2 to isolate ๐ฅ.
2
This idea of using inverses to solve equations continues when solving
logarithmic equations; we need to use the inverse of a logarithm in order
to solve a logarithmic equation, and that means converting to exponential
form.
Example 1 : Solve the logarithmic equation log
2
= โ 5 by converting
to exponential form. Simplify your answer completely.
To solve a logarithmic equation, convert to exponential form. Remember
that a logarithm is simply an exponent, so anything equal to a logarithm is
also an exponent. That means that if log
2
= โ 5 , then โ 5 is an
exponent.
log
2
= โ 5 converts to ๐ฅ = 2
โ 5
because โ 5 is an exponent
5
Example 2 : Solve each of the following equations by converting to
exponential form, and simplify your answers completely.
a. log
8
= 2 b. log
= โ 2 c. ln
Once again, to solve each logarithmic equation convert to exponential
form. Again, a logarithm is simply an exponent, so anything equal to
a logarithm is also an exponent.
log
8
= 2 log
10
= โ 2 log
๐
converts to converts to converts to
2
โ 2
0
๐
๐๐๐
Being able to convert from logarithmic form to exponential form is crucial
when solving logarithmic equations. Keep in mind that when converting
from one form to the other, THE BASE DOES NOT CHANGE. Base ๐
in one form is base ๐ in the other form; we simply switch the inputs and
outputs because logarithms and exponentials are inverses.
Example 3 : Solve each of the following equations by converting to
exponential form, and simplify your answers completely. DO NOT
APPROXIMATE, LEAVE ANSWERS IN EXACT FORM,.
a. log
27
1
3
b. log
= โ 4 c. ln
c. log (
5 ๐ฅ+ 1
2 ๐ฅโ 3
) = 2 d. log
27
2
3
d. ๐
log
๐
โ
2
3
โ 1
1
27
2
3
1
๐
1
( โ
27
3
)
2
1
( 3 )
2
1
9
1
9
๐๐
๐
e. log
4
๐ฅ+ 1
3 ๐ฅโ 2
1
2
f. ln
f.
๐ฅ+ 1
3 ๐ฅโ 2
โ
1
2
15
๐ฅ
๐ฅ
4
โ 1
๐ฅ+ 1
3 ๐ฅโ 2
1
4
1
2
15
๐ฅ
๐ฅ
4
1
8
๐ฅ+ 1
3 ๐ฅโ 2
1
โ 4
2
๐ฅ+ 1
3 ๐ฅโ 2
1
2
2
g. ln
= 1 h. log
2
h. A
log
๐
15
๐ฅ
๐ฅ
4
โ 1
1
15
๐ฅ
๐ฅ
4
1
8
2
i. ln( 4 ๐ฅ โ 29 ) = โ 1 j. log
8
15
๐ฅ
๐ฅ
4
j. H
log
๐
15
๐ฅ
๐ฅ
4
โ 1
โ 1
15
๐ฅ
๐ฅ
4
1
8
1
๐
15
๐ฅ
๐ฅ
4
1
8
2
2
2
๐๐
๐
