SOLUTION: Solve for b: logb8=-3

Question 240762: Solve for b:
logb8=-3
Found 2 solutions by jsmallt9, JimboP1977:
Answer by jsmallt9(3759) (Show Source): You can put this solution on YOUR website!
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 is equivalent to .

Rewriting your equation in exponential form we get:

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)

which results in:

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:

or

Since 1/3 as an exponent means "cube root of" and the cube root of 8 is 2:

Since -1 as an exponent means "reciprocal of" and the reciprocal of 2 is 1/2:

Answer by JimboP1977(311) (Show Source): You can put this solution on YOUR website!
b^(-3) = 8
(b^(-3))^(-1/3) = 8^-1/3
b^(-3*-1/3) = 8^-1/3
b^1 = 0.5
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