Log-ln

[stewja2]

:?:
Okay, I am sure this is simple; however, I am stuck!!!
I can't even begin to start working this problem.. Could someone please point me in the right direction....

ln(25x)-ln(x-1)=2

Where do I start?????

Any advice would be greatly appreciated!!

Jennifer

 

[pka]

Do you know that \(\displaystyle \ln (a) = b\quad \mbox{if and only if} \quad e^b = a\)?
Therefore from your problem we have \(\displaystyle \frac{{25x}}{{x - 1}} = e^2\)

So solve for x.

 

[stewja2]

Okay, I actually had that part but I wasn't sure it is was correct....

From that point to actually calculate e^2 before I solve for x?

25x= e^2(x-1)
25x= 7.389056099(x-1)

Thanks,
Jennifer

 

[stewja2]

Okay, I did that... Solved for x and came up with -.4196, however, unless I am confused it can't be a negative answer... Can it?

Thanks again,
Jennifer

 

[pka]

You are correct!
Any solution must be in the domain: x>1.