Equation

**joshirabi** · August 22nd 2006, 09:13 PM

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

**CaptainBlack** · August 22nd 2006, 10:44 PM

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

**ThePerfectHacker** · August 23rd 2006, 06:41 AM

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,
$x=\frac{2}{17}$

**joshirabi** · August 23rd 2006, 07:18 AM

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

**ThePerfectHacker** · August 23rd 2006, 07:33 AM

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,
$x=\frac{\log 72\cdot \log 9/8}{2}$

**Soroban** · August 23rd 2006, 08:25 PM

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

**ThePerfectHacker** · August 24th 2006, 11:34 AM

Originally Posted by Soroban

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

That is correct.

**Glaysher** · August 25th 2006, 01:44 AM

Alternatively written

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

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

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

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

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

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

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

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

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

**ThePerfectHacker** · August 25th 2006, 04:21 AM

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.

**Glaysher** · August 25th 2006, 04:32 AM

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.

**ThePerfectHacker** · August 25th 2006, 04:38 AM

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.

**Glaysher** · August 25th 2006, 04:43 AM

That makes a lot more sense.

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