Question 240762
When the variable is in the argument or the base of a logarithm, you solve for that variable by rewriting the equation in exponential form. This is done by using the fact that {{{log(a, (p)) = q}}} is equivalent to {{{a^q = p}}}.<br>
Rewriting your equation in exponential form we get:
{{{b^(-3) = 8}}}
Now we have an equation we can solve. There are several ways we could solve this. One would be to raise both sides to the -1/3 power. (You'll see wht when we're done)
{{{(b^(-3))^(-1/3) = 8^(-1/3)}}}
which results in:
{{{b =  8^(-1/3)}}}
We have been trying to solve for b and, as you can see, we have been able to do this in one step (raising to the -1/3 power). All that is left is to simpilfy the right side. To do this by hand it helps to factor the exponent:
{{{b =  8^((1/3)*(-1))}}}
or
{{{b = (8^(1/3))^(-1)}}}
Since 1/3 as an exponent means "cube root of" and the cube root of 8 is 2:
{{{b = 2^(-1)}}}
Since -1 as an exponent means "reciprocal of" and the reciprocal of 2 is 1/2:
{{{b = 1/2}}}