SOLUTION: Evaluate the log expression: {{{8^((log((base8))20))}}} So I know that log(base a)(b^r) = r log (base) b But I get lost on how to continue to break it down further from that it

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Evaluate the log expression: {{{8^((log((base8))20))}}} So I know that log(base a)(b^r) = r log (base) b But I get lost on how to continue to break it down further from that it      Log On


   

Question 914189: Evaluate the log expression: 8%5E%28%28log%28%28base8%29%2920%29%29
So I know that log(base a)(b^r) = r log (base) b
But I get lost on how to continue to break it down further from that it wouldn't be log(base 8) (20^8)?
Please help me figure this out
Found 3 solutions by josgarithmetic, MathLover1, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
INVERSE FUNCTIONS...
Do you recognize their use or occurrence in the expression?

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
8%5E%28%28log%288%2C20%29%29%29


=8^(log(20)/log(8))


=8^(log(2^2*5)/log(2^3))

=8^((2log(2)+log(5))/(3log(2))) ......plug in values

log%282%29=0.30102
log%285%29=0.69897

=8^((2(0.30102)+(0.69897))/(3(0.30102)))

=8^((0.60204+0.69897)/0.90306))

=8^(1.30101/0.90306)

=8%5E1.44066

=20.0007 round

=20


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Evaluate the log expression: 8%5E%28%28log%28%28base8%29%2920%29%29
So I know that log(base a)(b^r) = r log (base) b
But I get lost on how to continue to break it down further from that it wouldn't be log(base 8) (20^8)?
Please help me figure this out


Let 8%5E%28log+%288%2C+20%29%29 = x

log+%288%2C+x%29+=+log+%288%2C+20%29 -------Converting to LOGARITHMIC FORM

Thus, highlight_green%28highlight_green%28x_or_8%5E%28log+%288%2C+20%29%29+=+20%29%29