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

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The discussion revolves around solving the equation 8^x + 8^(-x) given that 4^x + 4^(-x) = 8. Participants attempted substitutions, particularly using y = 4^x, but faced challenges with real roots in the resulting quadratic equation. By expressing the terms in base 2, they derived that (y + 1/y) = √10, leading to the calculation of y^3 + 1/y^3. Ultimately, the computed result of 7√10 does not match any provided answer choices, suggesting a possible typo in the problem statement or that the correct answer is missing.
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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.
 
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Monoxdifly said:
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.
 
kaliprasad said:
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...
 
Monoxdifly said:
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.
 
Ah, okay then. Thanks to both of you.
 
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