Solve 8^x+8^(-x)=? When 4^x+4^(-x)=8

[Monoxdifly]

If $$4^x+4^{-x}=8$$, then $$8^x+8^{-x}=?$$
A. 14
B. 15
C. 16
D. 17
E. 18

What should I do? I tried substituting $$y = 4^x$$ but it didn't help since the quadratic equation formed didn't have real roots.

 

[kaliprasad]

If $$4^x+4^{-x}=8$$, then $$8^x+8^{-x}=?$$
A. 14
B. 15
C. 16
D. 17
E. 18

What should I do? I tried substituting $$y = 4^x$$ but it didn't help since the quadratic equation formed didn't have real roots.

we know $4=2^2$ and $8= 2^3$ so put $2^x = y$
we get $y^2 + \frac{1}{y^2} = 8\cdots(1)$
we need to find $y^3+ \frac{1}{y^3}$
from (1) we get
$(y + \frac{1}{y})^2 -2 = 8$
or $(y + \frac{1}{y}) = \sqrt{10}$

now you should be able to proceed.

 

[Monoxdifly]

we know $4=2^2$ and $8= 2^3$ so put $2^x = y$
we get $y^2 + \frac{1}{y^2} = 8\cdots(1)$
we need to find $y^3+ \frac{1}{y^3}$
from (1) we get
$(y + \frac{1}{y})^2 -2 = 8$
or $(y + \frac{1}{y}) = \sqrt{10}$

now you should be able to proceed.

Well...
$$(y + \frac{1}{y})^3 = (\sqrt{10})^3$$
$$y^3+3y^2(\frac{1}{y})+3y(\frac{1}{y^2})+\frac{1}{y^3}=10\sqrt{10}$$
$$y^3+\frac{1}{y^3}+3y+3(\frac1y)=10\sqrt{10}$$
$$y^3+\frac{1}{y^3}+3(y+\frac1y)=10\sqrt{10}$$
$$y^3+\frac{1}{y^3}+3\sqrt{10}=10\sqrt{10}$$
$$y^3+\frac{1}{y^3}=7\sqrt{10}$$
Not in the options...

 

[I like Serena]

Well...
$$(y + \frac{1}{y})^3 = (\sqrt{10})^3$$
$$y^3+3y^2(\frac{1}{y})+3y(\frac{1}{y^2})+\frac{1}{y^3}=10\sqrt{10}$$
$$y^3+\frac{1}{y^3}+3y+3(\frac1y)=10\sqrt{10}$$
$$y^3+\frac{1}{y^3}+3(y+\frac1y)=10\sqrt{10}$$
$$y^3+\frac{1}{y^3}+3\sqrt{10}=10\sqrt{10}$$
$$y^3+\frac{1}{y^3}=7\sqrt{10}$$
Not in the options...

Wolfram confirms that your solution is correct.
So it appears there is either a typo in the problem statement, or the correct answer is indeed not listed.

 

[Monoxdifly]

Ah, okay then. Thanks to both of you.