solve it pls
8^2x+1 = 9^2X-1
Please use brackets to resolve the ambiguity in what the exponents are,Originally Posted by joshirabi
also resolve the doubt about the intended identity of x and X.
(I can guess what the question is supposed to be but don't see why the
helpers here should have to reconstruct the problem).
RonL
Okay,Originally Posted by joshirabi
$\displaystyle \log 8^{2x+1}=\log 9^{2x-1}$
Thus,
$\displaystyle (2x+1)\log 8=(2x-1)\log 9$
$\displaystyle 2x\log 8+\log 8=2x\log 9-\log 9$
$\displaystyle 2x\log 8-2x\log 9=-\log 8-\log 9$
$\displaystyle 2x\log 9-2x\log 8=\log 9+\log 8$
$\displaystyle 2x(\log 9 - \log 8)=\log 9 +\log 8$
$\displaystyle 2x=\frac{\log 9+\log 8}{\log 9-\log 8}$
But,
$\displaystyle \log 9+\log 8 =\log (72)$
$\displaystyle \log 9 -\log 8=\log (9/8)$
Thus,
$\displaystyle x=\frac{\log 72\cdot \log 9/8}{2}$
Alternatively written
$\displaystyle 8^{2x + 1} = 9^{2x - 1}$
$\displaystyle e^{(\ln 8)^{2x + 1}} = e^{(\ln 9)^{2x - 1}}$
$\displaystyle e^{\ln 8(2x + 1)} = e^{\ln 9(2x - 1)}$
$\displaystyle \ln 8(2x + 1)} = \ln 9(2x - 1)$
$\displaystyle 2x \ln 8 + \ln 8 = 2x \ln 9 - \ln 9$
$\displaystyle 2x \ln 9 - 2x \ln 8 = \ln 8 + \ln 9$
$\displaystyle 2x(\ln 9 - \ln 8) = \ln 72$
$\displaystyle 2x = \frac{\ln 72}{\ln \frac{9}{8}}$
$\displaystyle x = \frac{\ln 72}{2\ln\frac{9}{8}}$
Don't know why I didn't just take logs of both sides would have saved a few steps but I don't understand why you putOriginally Posted by ThePerfectHacker
(3x + 7) + (4x - 9) + (5x - 6) = 12x - 8
at the start of your solution.
If it wasn't for that I wouldn't have bothered posting and I wanted something different. Nothing wrong with a different approach. Just go with the method you understand best.