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
2x = log_{5}16
2x =
2x 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}(3x) + log_{3}(x - 2) = 2
.
log_{3}(3x) + log_{3}(x - 2) = 2
log_{3}(3x(x - 2)) = 2
3^{2} = 3x(x - 2)
9 = 3x
^{2} - 6x
3x
^{2} - 6x - 9 = 0
3(x
^{2} - 2x - 3) = 0
3(x - 3)(x + 1) = 0
x = 3, - 1
Check:
Example 2: Solve for
x
:
2 log_{(2x+1)}(2x + 4) - log_{(2x+1)}4 = 2
.
2 log_{(2x+1)}(2x + 4) - log_{(2x+1)}4 = 2
log_{(2x+1)}(2x + 4)^{2} - log_{(2x+1)}4 = 2
log_{(2x+1)}
= 2
(2x + 1)^{2} =
(2x + 1)^{2} =
4x
^{2} +4x + 1 = x
^{2} + 4x + 4
3x
^{2} - 3 = 0
3(x
^{2} - 1) = 0
3(x + 1)(x - 1) = 1
x = 1, - 1
Check: