I have a log/exponential problem I need help with:

$\displaystyle 4^x + 6( 4 ^ {-x} ) = 5

$

So far the only thing i've thought of is to change 6 and 5 into a base of 4.

$\displaystyle 6 = 4^\frac {\log_6}{log_4} $

If u can help, thanks!

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- Oct 31st 2008, 04:11 PMUncle6Log/Exponential problem
I have a log/exponential problem I need help with:

$\displaystyle 4^x + 6( 4 ^ {-x} ) = 5

$

So far the only thing i've thought of is to change 6 and 5 into a base of 4.

$\displaystyle 6 = 4^\frac {\log_6}{log_4} $

If u can help, thanks!

- Oct 31st 2008, 04:15 PMJhevon
- Oct 31st 2008, 04:20 PMShyam
- Oct 31st 2008, 04:30 PMUncle6
Whoops, my bad, the question is actually has a $\displaystyle 4^{-x}$. Which is why I found this question to be so hard =S

- Oct 31st 2008, 04:37 PMJhevon
- Oct 31st 2008, 04:55 PMUncle6
uhh, I fixed it up there.

- Oct 31st 2008, 05:03 PMJhevon
ok, here's the trick

multiply through by $\displaystyle 4^x$, we get:

$\displaystyle (4^x)^2 + 6 = 5(4^x)$

or in other words,

$\displaystyle (4^x)^2 - 5(4^x) + 6 = 0$

now do you see what to do?

to make it a bit clearer, what if we made $\displaystyle a = 4^x$, then we have

$\displaystyle a^2 - 5a + 6 = 0$

now continue - Oct 31st 2008, 05:10 PMUncle6
WOW! that realy helps, thanks! I swear, I'd never see that trick..

- Oct 31st 2008, 05:26 PMJhevon
- Nov 1st 2008, 03:06 PMUncle6
x= 1/2 or around 0.792

- Nov 1st 2008, 04:39 PMShyam
$\displaystyle (4^x)^2 - 5(4^x) + 6 = 0$

$\displaystyle (4^x-2)(4^x-3)=0$

$\displaystyle 4^x=2$

$\displaystyle 2^{2x}=2^1$

2x=1

x=1/2

Also,

$\displaystyle 4^x=3$

$\displaystyle x \ln 4 = \ln 3$

$\displaystyle x = \frac{\ln 3}{\ln 4}=0.792$

x = 1/2, 0.792

your answers are right