SOLUTION: log25(base2).log8(base11).log11(base5)

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: log25(base2).log8(base11).log11(base5)      Log On


   

Question 276084: log25(base2).log8(base11).log11(base5)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%282%2C+%2825%29%29%2Alog%2811%2C+%288%29%29%2Alog%285%2C+%2811%29%29
An expression of logarithms is easier to work with if the bases are the same. Fortunately there is a formula for changing bases of logarithms:
log%28a%2C+%28p%29%29+=+log%28b%2C+%28p%29%29%2Flog%28b%2C+%28a%29%29
which expresses a base a logarithm in terms of base b logarithms. Since 25, the argument of the first log, is a power of 5 which is thebase of the third log and since the second log and third log both have 11's, it seems that converting the first two logs into the base of the third log, 5, holds promise:

Already we can see dividends. The second denominator and the third log cancel out:

leaving:
%28log%285%2C+%2825%29%29%2Flog%285%2C+%282%29%29%29%2Alog%285%2C+%288%29%29
And the first numerator is 2 (since 5%5E2+=+25):
%282%2Flog%285%2C+%282%29%29%29%2Alog%285%2C+%288%29%29
The next step will be clearer if I rearrange this a little:
2%28log%285%2C+%288%29%29%2Flog%285%2C+%282%29%29%29
Does the pattern of the fraction look familiar? If not, look at the base conversion formula. After a while I hope you recognize that this fraction is the result of converting log%282%2C+%288%29%29 into an expression of base 5 logs. So the fraction is simply log%282%2C+%288%29%29! (If you still don't see this, then just convert both the numerator and denominator logs into base 2 logs. It's more work this way but easier to see what's going on.) Replacing the fraction with thebase 2 log we get:
2log%282%2C+%288%29%29
Since 2%5E3+=+8, the remaining log is 3:
2%2A3
or
6