Monoxdifly said:If \(\displaystyle 4^x+4^{-x}=8\), then \(\displaystyle 8^x+8^{-x}=?\)
A. 14
B. 15
C. 16
D. 17
E. 18
What should I do? I tried substituting \(\displaystyle y = 4^x\) but it didn't help since the quadratic equation formed didn't have real roots.
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.
Monoxdifly said:Well...
\(\displaystyle (y + \frac{1}{y})^3 = (\sqrt{10})^3\)
\(\displaystyle y^3+3y^2(\frac{1}{y})+3y(\frac{1}{y^2})+\frac{1}{y^3}=10\sqrt{10}\)
\(\displaystyle y^3+\frac{1}{y^3}+3y+3(\frac1y)=10\sqrt{10}\)
\(\displaystyle y^3+\frac{1}{y^3}+3(y+\frac1y)=10\sqrt{10}\)
\(\displaystyle y^3+\frac{1}{y^3}+3\sqrt{10}=10\sqrt{10}\)
\(\displaystyle y^3+\frac{1}{y^3}=7\sqrt{10}\)
Not in the options...
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Yes, this equation can be solved without a calculator using algebraic methods. However, depending on the values of x and y, the solutions may involve irrational or complex numbers that may be difficult to calculate without a calculator.