# Simple log problem

1. Sep 18, 2004

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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

2. Sep 18, 2004

### 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

Last edited: Sep 18, 2004
3. Sep 18, 2004

### 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..

4. Sep 18, 2004

### Pyrrhus

Or simplest

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

Last edited: Sep 18, 2004
5. Sep 18, 2004

### Tide

Try $y = 4^x$.

6. Sep 18, 2004

### arildno

I would have said "Or simplest.." :shy: