How do you solve #log_x(1/8) = -3/2#?

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Answer 1 · 2017-06-29T05:19:31

The answer is #4#

1.) Convert to exponential form:
#x^(-3/2) = 1/8#

2.) Following rules of exponents, we know that #x^(-3/2) = 1/(x^(3/2)#

3.) Now our equation looks like this:

#1/(x^(3/2)) = 1/8#

4.) Cross-multiply and find that #x^(3/2) = 8#

5.) Convert to radical form:

#sqrt(x^3) = 8#

6.) Square both sides.

#x^3 = 64#

7.) Take the cube root of both sides.

#root3(x^3) = root3(64)#

#x = 4#