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.Com
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) (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
= 2
note that the 3 cancels into the 9 and x into x^2, leaving us with
= 2
The anti log of both sides, (10^x) on the calc)
= 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
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