- Thread starter
- #1

#### Monoxdifly

##### Well-known member

- Aug 6, 2015

- 265

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.

- Thread starter Monoxdifly
- Start date

- Thread starter
- #1

- Aug 6, 2015

- 265

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.

we know $4=2^2$ and $8= 2^3$ so put $2^x = y$

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.

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.

- Thread starter
- #3

- Aug 6, 2015

- 265

Well...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.

\(\displaystyle (y + \frac{1}{y})^3 = (\sqrt{10})^3\)

\(\displaystyle y^3+3y^2(\frac{1}{y})+3y(\frac{1}{y^2})+\frac{1}{y^3}=10\sqrt{10}\)

\(\displaystyle y^3+\frac{1}{y^3}+3y+3(\frac1y)=10\sqrt{10}\)

\(\displaystyle y^3+\frac{1}{y^3}+3(y+\frac1y)=10\sqrt{10}\)

\(\displaystyle y^3+\frac{1}{y^3}+3\sqrt{10}=10\sqrt{10}\)

\(\displaystyle y^3+\frac{1}{y^3}=7\sqrt{10}\)

Not in the options...

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- #4

Wolfram confirms that your solution is correct.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...

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

- Thread starter
- #5

- Aug 6, 2015

- 265

Ah, okay then. Thanks to both of you.