1. ## Solve the equation...

3^(x-2) = 4^(2-x)

I know how to do these when the bottom numbers can be divided by each other to get the top equations down to solve, but how can I do this when when the numbers on the bottom are different?

I was thinking to times the 3 and 4 to get,

12^(4x-8) = 12^(6-3x)

and then I could do it from there. But am I even going about it the way I have to?

thanks

2. Originally Posted by driverfan2008
3^(x-2) = 4^(2-x)

I know how to do these when the bottom numbers can be divided by each other to get the top equations down to solve, but how can I do this when when the numbers on the bottom are different?

I was thinking to times the 3 and 4 to get,

12^(4x-8) = 12^(6-3x)

and then I could do it from there. But am I even going about it the way I have to?

thanks
Do you know how to apply logarithms?

$3^{x-2}=4^{2-x}\implies (x-2)\log(3)=(2-x)\log(4)$.

Can you try taking it from here?

--Chris

3. Hello, driverfan2008!

$3^{x-2} \:= \:4^{2-x}$

With two different bases, I was going to suggest using logs . . .
. . then I saw something!

We have: . $3^{x-2} \;=\;4^{2-x}$

. . . . . . . . $3^{x-2} \;=\;4^{-(2-x)}$

. . . . . . . . $3^{x-2} \;=\;\frac{1}{4^{x-2}}$

. . . $(3^{x-2})(4^{x-2}) \;=\;1$

. . . . . . . $12^{x-2} \;=\;1$

. . . . . . . $12^{x-2} \;=\;12^0$

. . . . . . . $x - 2 \;=\;0$

. . . . . . . . . . $\boxed{x \;=\;2}$

4. Originally Posted by Soroban
Hello, driverfan2008!

With two different bases, I was going to suggest using logs . . .
. . then I saw something!

We have: . $3^{x-2} \;=\;4^{2-x}$

. . . . . . . . $3^{x-2} \;=\;4^{-(2-x)}$

. . . . . . . . $3^{x-2} \;=\;\frac{1}{4^{x-2}}$

. . . $(3^{x-2})(4^{x-2}) \;=\;1$

. . . . . . . $12^{x-2} \;=\;1$

. . . . . . . $12^{x-2} \;=\;12^0$

. . . . . . . $x - 2 \;=\;0$

. . . . . . . . . . $\boxed{x \;=\;2}$