SOLUTION: write a rule for log (a)^m. Then use the idea of powers of ten to argue that your rule works for any m and all values of a 0.
I suppose that log (am)^m = m log a, but how do i
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-> SOLUTION: write a rule for log (a)^m. Then use the idea of powers of ten to argue that your rule works for any m and all values of a 0.
I suppose that log (am)^m = m log a, but how do i
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Question 281235: write a rule for log (a)^m. Then use the idea of powers of ten to argue that your rule works for any m and all values of a 0.
I suppose that log (am)^m = m log a, but how do i use the idea of powers of ten to prove? Please help!
Note: this problem is meant to help me figure out the power property of log:
log(lower case b) (a^n) = n x log(lower case b)(a) Answer by dabanfield(803) (Show Source):
You can put this solution on YOUR website! write a rule for log (a)^m. Then use the idea of powers of ten to argue that your rule works for any m and all values of a 0.
I suppose that log (am)^m = m log a, but how do i use the idea of powers of ten to prove? Please help!
Note: this problem is meant to help me figure out the power property of log:
log(lower case b) (a^n) = n x log(lower case b)(a)
Show why it makes sense that the log (a)^n = n*log a.
Remember 10^n = 10*10*10...*10 (n 10's).
So we have:
1.) log 10^n = log (10*10*10...*10) (n 10's)
Using the log rule that says log (a*b) = log a + log b then 1.) above becomes:
log 10^n = log 10 + log 10 + log 10 +...+ log 10 (n log 10's) so
log 10^n = n*log 10
(