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]

<br /> \begin{multline*}<br /> \begin{split}<br /> &amp;log(AB)=log\ A+log\ B\\<br /> &amp;There\ isn&#039;t\ anything\ like\ this:\\<br /> &amp;log(A+B)=log(AB)\\<br /> &amp;4^x+4^{-x}=4^x+\frac{1}{4^x}=\frac{4^{2x}+1}{4^x}\\<br /> &amp;4^x*4^{-x}=4^{x-x}=1\\<br /> \end{split}<br /> \end{multline*}<br />

 

[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.