# Exponential equation

• Aug 2nd 2009, 04:25 AM
bael
Exponential equation
I've banged my head senseless on an exponential equation lately. I can't seem to solve however i try. Any help would be appreciated.

problem:

4^2x - 4^x = 8 * 4^(x+3)

solve for x

• Aug 2nd 2009, 04:55 AM
flyingsquirrel
Hi,
Quote:

Originally Posted by bael
problem:

4^2x - 4^x = 8 * 4^(x+3)

solve for x

This equation can be written $\displaystyle 4^{x}\left(4^x-1\right)=8\times 4^x\times 4^3$ and one can divide both sides by $\displaystyle 4^x$ since for all $\displaystyle x$, $\displaystyle 4^x\neq0$.

Can you take it from here ?
• Aug 2nd 2009, 04:59 AM
CaptainBlack
Quote:

Originally Posted by bael
I've banged my head senseless on an exponential equation lately. I can't seem to solve however i try. Any help would be appreciated.

problem:

4^2x - 4^x = 8 * 4^(x+3)

solve for x

Lets assume the equation is:

$\displaystyle 4^{2x}-4^x=8 \times 4^{x+3}$

or:

$\displaystyle 4^{2x}-4^x=8 \times 4^3 \times 4^{x}$

rearrange:

$\displaystyle (4^{x})^2-513\times 4^x=4^x(4^x-513)=0$

so either $\displaystyle 4^x=0$ which is impossible or:

$\displaystyle 4^x=513$

and you should be able to finish this yourself from that.

CB
• Aug 2nd 2009, 05:02 AM
artvandalay11
Quote:

Originally Posted by bael
I've banged my head senseless on an exponential equation lately. I can't seem to solve however i try. Any help would be appreciated.

problem:

4^2x - 4^x = 8 * 4^(x+3)

solve for x

$\displaystyle 4^{2x}-4^x=8*4^{x+3}$
$\displaystyle 4^x*4^x-4^x=8*4^x*4^3$
$\displaystyle 4^x(4^x-1)=8*4^x*4^3$ and since $\displaystyle 4^x\not=0$ at anypoint we can divide by it
$\displaystyle 4^x-1=8*64$