1. ## Solving with logs

How do you solve this problem leaving logs in there?

2. Originally Posted by Togechu64
How do you solve this problem leaving logs in there?

$\ln{6^{7x+1}}=\ln{7^{6x-1}}$

$(7x+1)\ln(6)=(6x-1)\ln(7)$

Now distribute the lns, then solve for x.

3. I tried that, but I'm not sure where to go after I distribute? :/

4. $(7x+1)\ln(6)=(6x-1)\ln(7)$

Put all logs on a side, and the $x$ terms on the other side (distribute the ln's, that is) :

$\frac{7x+1}{6x - 1} = \frac{\ln(7)}{\ln(6)}$

Say $a = \frac{\ln(7)}{\ln(6)}$ (for ease of manipulation and handling) :

$\frac{7x + 1}{6x - 1} = a$

Get rid of the fraction by multiplying each side by $(6x - 1)$.

$7x + 1 = a(6x - 1)$

Then expand and solve for $x$ (then substitute for $a$)

Does it help ?

5. yes it does, thank you very much!!