1. ## (8^x)+(8^(-x))

if (4^x)+(4^(-x))=7

(8^x)+(8^(-x))= ????

thanks

2. Hello, Singular!

This is a most unplesant problem . . . especially without LaTeX.

If .4^x + 4^{-x} .= .7, then find: .8^x + 8^{-x}
The given equation is: .2^{2x} + 2^{-2x} .= .7

Cube both sides: .[2^{2x} + 2^{-2x}]³ ,= ,

. . 2^{6x} + 3[2^{4x}][2^{-2x}] + 3[2^{2x}][2^{-4x}] + 2^{-6x} .= .343

. . (2³)^{2x} + 3[2^{2x}] + 3[2^{-2x}] + (2³)^{-2x} .= .343

. . 8^{2x} + 3[4^x + 4^{-x}] + 8^{2x} .= .343
. . . . . . . . . . .\__________/
. . . . . . . . . . . . .
This is 7

So we have: .8^{2x} + 21 + 8^{-2x} .= .343 . . 8^{2x} + 8^{-2x} .= .322

. . . . . . . . . (8^x)² + (8^{-x})² .= .322 . [1]

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We will take a slight detour now . . .

Consider the square of: 8^x + 8^{-x} .= .k

We have: .[8^x + 8^{-x}]² .= .(8^x)² + 2(8^x)(8^{-x}) + (8^{-x})² .= .

. . . . . . . . (8^x)² + 2 + (8^{x})² .= . . . (8^x)² + (8^{-x})² .= .k² - 2

Substitute into [1]: .k² - 2 .= .322 . . k² = 324 . . k = 18

Therefore: . 8^x + 8^{-x} .= .18

Is there a better way? . . . I hope so!

3. Originally Posted by Soroban
Hello, Singular!

This is a most unplesant problem . . . especially without LaTeX.

2^{2x} + 2^{-2x}=7
Why do you say that?
The problem does not say what this is.

4. If
2^{2x} + 2^{-2x}
is not equal to
4^x + 4^{-x}
then what should it be?

-Dan

5. Code:
8^{x}+8^{-x}=2^{3x}+2^{-3x}
=[2^{x}+2^{-x}][2^{2x}-1+2^{-2x}]
=[2^{x}+2^{-x}][4^{x}-1+4^{-x}]
=[2^{x}+2^{-x}][6]

Now find the value of

[2^{x}+2^{-x}]^{2}=[2^{2x}+2+2^{-2x}]
=9

Thus (3)(6)=18