# Log/Exponential problem

• Oct 31st 2008, 04:11 PM
Uncle6
Log/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 PM
Jhevon
Quote:

Originally Posted by Uncle6
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!

why would you do that?

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

$\displaystyle \Rightarrow 7(4^x) = 5$

$\displaystyle \Rightarrow 4^x = \frac 57$

can you finish?
• Oct 31st 2008, 04:20 PM
Shyam
Quote:

Originally Posted by Uncle6
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!

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

take common, $\displaystyle 4^x$

$\displaystyle 4^x(1+6)=5$

$\displaystyle 4^x=\frac{5}{7}$

Now, take natural log on both sides,

$\displaystyle \ln \;(4^x)=\ln \left(\frac{5}{7}\right)$

now simplify further.
• Oct 31st 2008, 04:30 PM
Uncle6
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 PM
Jhevon
Quote:

Originally Posted by Uncle6
Whoops, my bad, the question is actually has a $\displaystyle 4^{-x}$. Which is why I found this question to be so hard =S

which of the $\displaystyle 4^x$'s should be $\displaystyle 4^{-x}$??
• Oct 31st 2008, 04:55 PM
Uncle6
uhh, I fixed it up there.
• Oct 31st 2008, 05:03 PM
Jhevon
Quote:

Originally Posted by Uncle6
uhh, I fixed it up there.

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 PM
Uncle6
WOW! that realy helps, thanks! I swear, I'd never see that trick..
• Oct 31st 2008, 05:26 PM
Jhevon
Quote:

Originally Posted by Uncle6
WOW! that realy helps, thanks! I swear, I'd never see that trick..

tell us your solution when finished
• Nov 1st 2008, 03:06 PM
Uncle6
x= 1/2 or around 0.792
• Nov 1st 2008, 04:39 PM
Shyam
Quote:

Originally Posted by Uncle6
x= 1/2 or around 0.792

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

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

$\displaystyle 4^x=2$

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

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