SOLUTION: log_3(x) - log_9(x) = 2 First i changed the base, by doing a little basic reasoning. Suppose y = log_9(x) Then 9^y = x (3^2)^y = x 3^(2y) = x

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: log_3(x) - log_9(x) = 2 First i changed the base, by doing a little basic reasoning. Suppose y = log_9(x) Then 9^y = x (3^2)^y = x 3^(2y) = x       Log On


   

Question 70656: log_3(x) - log_9(x) = 2


First i changed the base, by doing a little basic
reasoning. Suppose
y = log_9(x)
Then
9^y = x
(3^2)^y = x
3^(2y) = x
2y = log_3(x)
y = log_3(x)/ 2
Now you have

log_3(x) - (1/2)log_3 = 2

but im stuck from here, please help me with solving trhe x? ironicly i know the answer is 81, but i came to that but doing trial and error log_3(79) - log_9(79) = 1.9877etc than log_3(81) - log_9(81) = 2. I am sure there is a much simplar way of dealing with this but i dont know how to deal with the x above.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
Log%5B3%5Dx+-Log%5B9%5Dx+=+2 Change the base of the first log from 3 to 9
Log%5B3%5Dx+=+%28Log%5B9%5Dx%29%2FLog%5B9%5D3
But Log%5B9%5D3+=+1%2F2 So now you have:
Log%5B3%5Dx+=+2Log%5B9%5Dx
2Log%5B9%5Dx-Log%5B9%5Dx+=+2 Subtract the logarithms.
Log%5B9%5Dx+=+2 Rewrite in exponential form..
9%5E2+=+x
x+=+81