Solving a Logarithmic EquationSolve
Quotient ruleProperty 4Multiply by x.
Subtract x.Divide by 6.Since we cannot take the logarithm of a nonpositivenumber, both and xmust
be positive here. If then this condition is satisfied.
CHECK Original equation
LetAdd.Quotient rule✓True
A true statement results, so the solution set is E NOW TRY
16 F.
log 2 7 = log 27
log 2
7
6
1
6log
2 7
log 2
7
6
- log 2
1
6
log
2 7
log 2 a x=^16.
1
6
+ 1 b - log 2
1
6
log
2 7
log 2 1 x+ 12 - log 2 x= log 2 7
x=
16 ,
x+ 1
1
6
= x
1 = 6 x
x+ 1 = 7 x
x+ 1
x
= 7
log 2
x+ 1
x
= log 2 7
log 2 1 x+ 12 - log 2 x= log 2 7
log 2 1 x+ 12 - log 2 x= log 2 7.
EXAMPLE 4
SECTION 10.6 Exponential and Logarithmic Equations; Further Applications 615
Transform the left
side to an expresssion
with only one
logarithm.This proposed
solution must
be checked.7
6
1
6=^76 ,^16 =^76 #^61 = 7
Solving a Logarithmic EquationSolve
Product rule
Write in exponential form.
Distributive property; multiply.
Standard form
Factor.or Zero-factor property
or Proposed solutions
The value must be rejected as a solution since it leads to the logarithm of a nega-
tive number in the original equation.
The left side is undefined.Check that the only solution is 25, so the solution set is 5256. NOW TRY
log 1 - 42 + log 1 - 4 - 212 = 2
- 4
x= 25 x=- 4
x- 25 = 0 x + 4 = 0
1 x- 2521 x+ 42 = 0
x^2 - 21 x- 100 = 0
x^2 - 21 x= 100
x 1 x- 212 = 102
log x 1 x- 212 = 2
log x+log 1 x- 212 = 2
log x+ log 1 x- 212 = 2.
EXAMPLE 5
The base
is 10.NOW TRY
EXERCISE 4
Solve.
=log 4 10log 4 12 x+ 132 - log 4 1 x+ 12
NOW TRY
EXERCISE 5
Solve.
log 4 1 x+ 22 +log 4 2 x= 2NOW TRY ANSWERS
4.E^38 F 5. 526