Simple log problem

[courtrigrad]

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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 dont see what I am doing wrong. I used the Product Rule. Any help would be greatly appreciated.

Thanks

[jcsd]

log (4^x + 4^-x) is not equivalent to log(4^x) + log(4^-x).

try using the fact that 4^x = 4^(2x)*4^-x

[arildno]

1.The product rule of logs says:
log(ab)=log(a)+log(b)
This is not what you've been using.
2. You should write, for example:
4^{x}=e^{xln(4)},4^{-x}=e^{-xln(4)}
Try to rewrite your original problem with a hyperbolic function..

[Pyrrhus]

Or simplest

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

Edit: To please arildno :rofl:

[Tide]

Try y = 4^x.

[arildno]

Or simply

(4^x)(4^x) + (4^-^x)(4^x) = \frac{5}{2}(4^x)
I would have said "Or simplest.." :shy: :redface: :cry: