# Thread: Equation

1. ## Equation

solve it pls
8^2x+1 = 9^2X-1

2. Originally Posted by joshirabi
solve it pls
8^2x+1 = 9^2X-1
Please use brackets to resolve the ambiguity in what the exponents are,
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

3. Originally Posted by joshirabi
solve it pls
8^2x+1 = 9^2X-1
8^2=64
9^2=81
Thus,
$64x+1=81x-1$
Subtract 64x from both sides,
$1=17x-1$
Add 1 to both sides,
$2=17x$
Thus,
$x=\frac{2}{17}$

4. ## re equation

solve this equation pls
8^(2x+1) = 9^(2X-1)
where (2x+1) is power of 8 and (2x-1) is power of 9.

5. Originally Posted by joshirabi
solve this equation pls
8^(2x+1) = 9^(2X-1)
where (2x+1) is power of 8 and (2x-1) is power of 9.
Okay,
$\log 8^{2x+1}=\log 9^{2x-1}$
Thus,
$(2x+1)\log 8=(2x-1)\log 9$
$2x\log 8+\log 8=2x\log 9-\log 9$
$2x\log 8-2x\log 9=-\log 8-\log 9$
$2x\log 9-2x\log 8=\log 9+\log 8$
$2x(\log 9 - \log 8)=\log 9 +\log 8$

$2x=\frac{\log 9+\log 8}{\log 9-\log 8}$
But,
$\log 9+\log 8 =\log (72)$
$\log 9 -\log 8=\log (9/8)$
Thus,
$x=\frac{\log 72\cdot \log 9/8}{2}$

6. Of course, The Perfect Hacker meant: . $x \:=\:\frac{\log72}{2\log\frac{9}{8}}$

7. Originally Posted by Soroban
Of course, The Perfect Hacker meant: . $x \:=\:\frac{\log72}{2\log\frac{9}{8}}$

That is correct.

8. Alternatively written

$8^{2x + 1} = 9^{2x - 1}$

$e^{(\ln 8)^{2x + 1}} = e^{(\ln 9)^{2x - 1}}$

$e^{\ln 8(2x + 1)} = e^{\ln 9(2x - 1)}$

$\ln 8(2x + 1)} = \ln 9(2x - 1)$

$2x \ln 8 + \ln 8 = 2x \ln 9 - \ln 9$

$2x \ln 9 - 2x \ln 8 = \ln 8 + \ln 9$

$2x(\ln 9 - \ln 8) = \ln 72$

$2x = \frac{\ln 72}{\ln \frac{9}{8}}$

$x = \frac{\ln 72}{2\ln\frac{9}{8}}$

9. Originally Posted by Glaysher
Alternatively written...
I know I have a problem of being sometimes complicated. But I would not post something like that. It will confuse the student.

10. Originally Posted by ThePerfectHacker
I know I have a problem of being sometimes complicated. But I would not post something like that. It will confuse the student.
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 put

(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.

11. Originally Posted by Glaysher
but I don't understand why you put

(3x + 7) + (4x - 9) + (5x - 6) = 12x - 8

at the start of your solution.
LaTex error.

12. That makes a lot more sense.