SOLUTION: Simple logarithm problem-- Could someone please help me? I've got a test tomorrow and my professor did not have time to explain. I'm stuck! log3x^2=2+log9x Thank you!

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Simple logarithm problem-- Could someone please help me? I've got a test tomorrow and my professor did not have time to explain. I'm stuck! log3x^2=2+log9x Thank you!      Log On


   

Question 796118: Simple logarithm problem--
Could someone please help me? I've got a test tomorrow and my professor did not have time to explain. I'm stuck!
log3x^2=2+log9x
Thank you!
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
log3x^2=2+log9x
Assume it is:
log(3x^2) = 2 + log(9x)
Subtract log(9x) from both sides
log(3x^2) - log(9x) = 2
Subtracting logs is the same as divide so we can write this
log%28%28%283x%5E2%29%2F%289x%29%29%29 = 2
note that the 3 cancels into the 9 and x into x^2, leaving us with
log%28%28x%2F3%29%29 = 2
The anti log of both sides, (10^x) on the calc)
x%2F3 = 100
x = 3*100
x = 300
:
:
See if that checks out
Enter in your calc: log(3*300^2) = 5.431..
Enter: 2 + log(9*300) = 5.431.. equality reigns