SOLUTION: Given that {{{u = log(base9)x}}}, find in terms of u, i) {{{log(base3)x}}} ii) {{{log(base9)27x}}} iii) {{{log(base2)81}}} *Please answer as soon as possible bro. :) =)

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Given that {{{u = log(base9)x}}}, find in terms of u, i) {{{log(base3)x}}} ii) {{{log(base9)27x}}} iii) {{{log(base2)81}}} *Please answer as soon as possible bro. :) =)      Log On


   

Question 466508: Given that u+=+log%28base9%29x, find in terms of u,
i) log%28base3%29x
ii) log%28base9%2927x
iii) log%28base2%2981
*Please answer as soon as possible bro. :) =)
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Given that u+=+log%28base9%29x, find in terms of u,
i) log%28base3%29x
ii) log%28base9%2927x
iii) log%28base2%2981
...
log9(x)=u
exponential form:9^u=x
..
log3(x)
exponential form:(3^2)^u=3^2u=9^u=x
log3(x)=2u
..
log9(27x)
log9(x)+log9(27)
log9(x)+log9(9)+Log9(3)
log9(27x)=u+1+1/2=u+3/2
..
log2(81)
sorry, I don't know how to do this one.
Let me know when you find the right ans.