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

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Discussion Overview

The discussion centers around the mathematical problem of determining the value of $$8^x + 8^{-x}$$ given that $$4^x + 4^{-x} = 8$$. Participants explore various approaches to solve this problem, including substitutions and algebraic manipulations.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Debate/contested

Main Points Raised

  • One participant suggests substituting $$y = 4^x$$ but finds that the resulting quadratic equation does not yield real roots.
  • Another participant proposes using the relationship between powers of 2, leading to the equation $$y^2 + \frac{1}{y^2} = 8$$ and aims to find $$y^3 + \frac{1}{y^3}$$.
  • From the derived equation, it is noted that $$y + \frac{1}{y} = \sqrt{10}$$, which is used to further manipulate the expression for $$y^3 + \frac{1}{y^3}$$.
  • A calculation is presented that results in $$y^3 + \frac{1}{y^3} = 7\sqrt{10}$$, which does not match any of the provided answer options.
  • A later reply mentions that a computational tool confirms the derived solution, suggesting a possible typo in the problem statement or that the correct answer is not listed among the options.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct answer, as the derived solution does not align with the provided options. There is acknowledgment of a potential error in the problem statement.

Contextual Notes

The discussion highlights the limitations of the problem's formulation, particularly regarding the validity of the answer choices in relation to the derived mathematical expressions.

Monoxdifly
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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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