1. ## Help Simplifying Logarithms

I have a problem here involving natural logs, please show your work so I can learn if I'm making a mistake.

Simplify the following expression as completely as possible:
5x + ln (3x) - ln (9x^2) + ln (3) - ln (x) + ln (x^2) -4x

I know I can put the 2 exponential numbers in front of their natural logs ex. 2 ln (9x), 2 ln (x), but I'm having trouble deciding where to divide/multiply/etc.

2. Originally Posted by Farnsworth
I have a problem here involving natural logs, please show your work so I can learn if I'm making a mistake.

Simplify the following expression as completely as possible:

$5x + ln (3x) - ln (9x^2) + ln (3) - ln (x) + ln (x^2) -4x$

I know I can put the 2 exponential numbers in front of their natural logs ex. 2 ln (9x), 2 ln (x), but I'm having trouble deciding where to divide/multiply/etc.
You can evaluate ln(3) with your calculator, best not to though!.
5x-4x=x.

You need the division law of logarithms for the subtraction of logarithms

$ln(3x)-ln(9x^2)\ and\ ln(x^2)-ln(x)$

or

$ln(3x)-ln(x)\ and\ ln(x^2)-ln(9x^2)$

Use the logarithm law

$log_e(a)-log_e(b)=log_e\frac{a}{b}$

3. First you have a mistake: $\ln(9x^2)=2\ln(3x)$.
Do you see why?

4. so going off ln(3x) - ln(x) and ln(x^2) - ln (9x^2) the answer would come out as ln(3) - ln(9), then I could add the former ln(3) and x to get

2 ln(3) - ln(9) + x which would cancel out to x?

Man, if that's right I'm going to be so glad I asked you guys. *hopes*

5. Originally Posted by Farnsworth
so going off ln(3x) - ln(x) and ln(x^2) - ln (9x^2) the answer would come out as ln(3) - ln(9)
I'm getting $\ln\,3 + \ln \left( \frac{1}{9} \right)$ here.
Which equals
$\ln\,3 -2 \ln\,3 = -\ln\,3$

6. Originally Posted by eumyang
I'm getting $\ln\,3 + \ln \left( \frac{1}{9} \right)$ here.
Which equals
$\ln\,3 -2 \ln\,3 = -\ln\,3$
Yeah, but ln (3) + ln (1/9) and ln(3) - ln(9) both come out as the answer to -ln(3), so adding the first ln(3) would leave me with x, right?