Author Topic: exponential equation (Read 775 times) Tweet Share

/0 · « **on:** December 22, 2008, 03:32:48 pm »

A friend asked me this, and I can't think of how to do it, blah

$3^x = 4^y = 12^z$

Prove that $z = \frac{xy}{x+y}$

thanx

humph · « **Reply #1 on:** December 22, 2008, 04:06:12 pm »

Use the fact that $12^z = 3^z 4^z$ and take logs, such that $x = \log_3 (3^z 4^z) = \log_3 (3^z) + \log_3 (4^z) = z + z\log_3 4 = z(1 + \log_3 4)$ . Similarly $y = z(1 + \log_4 3)$ . Now sub these values in for $\frac{xy}{x+y}$ , and use the fact that $\log_3 4 \log_4 3 = 1$ .

/0 · « **Reply #2 on:** December 22, 2008, 04:10:25 pm »

Thanks humph, but I was wondering... is there a way to do this without using the answer?

NE2000 · « **Reply #3 on:** December 22, 2008, 04:50:29 pm »

$3^x=4^y=12^z=3^z 4^z$

$3^x=3^z 4^z$
$3^{x-z}=4^z$
$\log _4 3^{x-z} = z$
$(x-z) \log _4 3 = z$

Do the same with the other one and you get

$(y-z) \log _3 4 = z$

$z^2 = ((x-z) \log _4 3)((y-z) \log _3 4)$
$z^2 = (x-z)(y-z)$

as $\log _4 3 \log _3 4 = 1$

Hence
$z^2 = xy - zy - zx + z^2$
$0 = xy - z(y+x)$
$z = \frac {xy}{x+y}$

I guess others would have a better way

/0 · « **Reply #4 on:** December 22, 2008, 04:53:28 pm »

Oh i see, thanks

I was trying to do it with logs to an unknown base.

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