To solve an equation containing a variable exponent, isolate the exponential quantity. Then take a logarithm, to the base of the exponent, of both sides.

*Example 1*: Solve for *x*: 3^{x} = 15.

3^{x} = 15

log_{3}3^{x} = log_{3}15*x* = log_{3}15*x* = *x* 2.465

*Example 2*: Solve for *x*: 4·5^{2x} = 64.

4·5^{2x} = 64

5^{2x} = 16

log_{5}5^{2x} = log_{5}16

2*x* = log_{5}16

2*x* =

2*x* 1.723*x* 0.861

To solve an equation containing a logarithm, use the properties of logarithms to combine the logarithmic expressions into one expression. Then convert to exponential form and evaluate. Check the solution(s) and eliminate any extraneous solutions--recall that we cannot take the logarithm of a negative number.

*Example 1*: Solve for *x*: log_{3}(3*x*) + log_{3}(*x* - 2) = 2.

log_{3}(3*x*) + log_{3}(*x* - 2) = 2

log_{3}(3*x*(*x* - 2)) = 2

3^{2} = 3*x*(*x* - 2)

9 = 3*x*^{2} - 6*x*

3*x*^{2} - 6*x* - 9 = 0

3(*x*^{2} - 2*x* - 3) = 0

3(*x* - 3)(*x* + 1) = 0*x* = 3, - 1

Check:

*x*= 3: log_{3}(3·3) + log_{3}1 = 2 + 0 = 2.*x*= 3 is a solution.*x*= - 1: log_{3}(3· -1) + log_{3}(- 1 - 2) = log_{3}(- 3) + log_{3}(- 3) does not exist.*x*= - 1 is not a solution.

*Example 2*: Solve for *x*: 2 log_{(2x+1)}(2*x* + 4) - log_{(2x+1)}4 = 2.

2 log_{(2x+1)}(2*x* + 4) - log_{(2x+1)}4 = 2

log_{(2x+1)}(2*x* + 4)^{2} - log_{(2x+1)}4 = 2

log_{(2x+1)} = 2

(2*x* + 1)^{2} =

(2*x* + 1)^{2} =

4*x*^{2} +4*x* + 1 = *x*^{2} + 4*x* + 4

3*x*^{2} - 3 = 0

3(*x*^{2} - 1) = 0

3(*x* + 1)(*x* - 1) = 1*x* = 1, - 1

Check:

*x*= 1: 2 log_{3}6 - log_{3}4 = log_{3}6^{2}- log_{3}4 = log_{3}= log_{3}9 = 2.*x*= 1 is a solution.*x*= - 1: 2 log_{-1}2 - log_{-1}4 does not exist (the base cannot be negative).

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