Log / Exponential problem: solving 4^x + 6(4^(-x)) = 5

[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!

 

[Deleted member 4993]

Re: Log/Exponential problem

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

$$4^x + 6( 4 ^ x ) = 5$$
So far the only thing i've thought of is to change 6 and 5 into a base of 4.

$$6 = 4^\frac {\log_6}{log_4}$$
If u can help, thanks!

[sup:1xqa3l17][/sup:1xqa3l17]

$\displaystyle 4^{x} + 6\cdot 4^{-x} \, = \, 5$

substitute:

$\displaystyle u \, = \, 4^x$

then

$\displaystyle u \, + \, \frac{6}{u} \, = \, 5$

Now you have a quadratic equation. Solve for 'u' and then solve for 'x' using log.

 

[Uncle6]

Re: Log/Exponential problem

My bad, the question actually has a $$4^{-x}$$ in it. That's why it's hard and I can't solve! help plz =S

 

[Deleted member 4993]

Re: Log/Exponential problem

My bad, the question actually has a $$4^{-x}$$ in it. That's why it's hard and I can't solve! help plz =S

I fixed it above....

 

[Uncle6]

Re: Log/Exponential problem

Thanks for help so far, but I still have difficulty understanding. Doesn't a quadratic equation have to be in the form of ax^2+bx+c=o ?

I'm not really sure how to solve for u=4^x . do I change it into log form? log_{4} u

 

[Uncle6]

Re: Log/Exponential problem

I think I got it. just multply both sides by 4^x. Thanks.

 

[mmm4444bot]

Re: Log/Exponential problem

I think I got it. just multply both sides by 4^x

$\displaystyle 4^x \cdot 4^{x} + 6\cdot 4^x \cdot 4^{-x} \, = \, 5 \cdot 4^x$

$\displaystyle 4^{2x} - 5 \cdot 4^x + 6 = 0$

$\displaystyle u \;=\; 4^x$

$\displaystyle u^2 - 5 \cdot u + 6 = 0$

$\displaystyle u + \frac{6}{u} = 5$