log3^(x+2)-log4^(x-3)=2

Question

anonymous · Answer

log a - log b = log (a/b)

anonymous · Answer

The two bases are different though. Could you explain it a little more in depth?

anonymous · Answer

Do you mean $$\log _{3} (x+2) - \log _{4} (x-3) = 2$$?

anonymous · Answer

Yes

anonymous · Answer

$$\log _{b} a = (\log a) / (\log b)$$

anonymous · Answer

So I take log3^(x+2) and divide it by log4^(x-3)...correct?

anonymous · Answer

No.  $$\log _{3} (x+2) = \log (x+2) / \log 3$$  The same operation can be applied to the other term.  But after that it's a bit tricky.  Are you sure you don't mean$$\log 3^{x+2} - \log 4^{x-3} = 2$$

anonymous · Answer

According to the worksheet, the powers are enclosed with parentheses.

anonymous · Answer

Since you said "powers" I'm going to assume that 3 and 4 are not the bases of the logarithms.  Hence my very first comment holds.  You can use that to combine the two logs into one, then take 10 to the power on each side of the equation and get rid of the log.

anonymous · Answer

Ok that makes sense.

anonymous · Answer

I had to do quite a bit of gymnastics to get it the rest of the way, though.  Post again if you get stuck.

anonymous · Answer

Ok thanks.