Solving a Simple Log Problem: 4^x + 4^-x = 5/2

[courtrigrad]

Hello,

I need help with this problem:


4^x + 4^-x = 5/2

My Solution: (Assume we use log with base 4)

log ( 4^x + 4^-x) = log(5/2)

= log (4^0) = log (5/2) ?


I don't see what I am doing wrong. I used the Product Rule. Any help would be greatly appreciated.

Thanks

 

[Pyrrhus]

How about this

$$(4^x)(4^x) + (4^-^x)(4 ^x) = \frac{5}{2}(4^x)$$

Maybe i should arrange

$$(4^x)^2 + 1 = \frac{5}{2}(4^x)$$

$$x= 4^x$$

$$x^2 + 1 = \frac{5}{2}x$$

 

[Leong]

$$\begin{multline*} \begin{split} &log(AB)=log\ A+log\ B\\ &There\ isn't\ anything\ like\ this:\\ &log(A+B)=log(AB)\\ &4^x+4^{-x}=4^x+\frac{1}{4^x}=\frac{4^{2x}+1}{4^x}\\ &4^x*4^{-x}=4^{x-x}=1\\ \end{split} \end{multline*}$$

 

[recon]

How about this

$$(4^x)(4^x) + (4^-^x)(4 ^x) = \frac{5}{2}(4^x)$$

Maybe i should arrange

$$(4^x)^2 + 1 = \frac{5}{2}(4^x)$$

$$x= 4^x$$

$$x^2 + 1 = \frac{5}{2}x$$

$$x= 4^x$$ should not be used. I see where you are going here. But this is wrong. Instead we should assign it to a different variable like $$y= 4^x$$. So,

$$y^2 + 1 = \frac{5}{2}y$$

 

[Pyrrhus]

It's true, i was just reminding him of Ax^2 + Bx + C, but thanks for pointing it out.