Equation

[123]

Given equation: \(\displaystyle x\cdot \sqrt{x}\cdot \sqrt[4]{x}\cdot \sqrt[8]{x}\cdot \sqrt[16]{x}\cdot ...=256\)
1) Show, that \(\displaystyle x\cdot \sqrt{x}\cdot \sqrt[4]{x}\cdot \sqrt[8]{x}\cdot \sqrt[16]{x}\cdot ...=x^2\)
2) Solve the equation.

 

[galactus]

\(\displaystyle x\cdot\sqrt{x}\cdot\sqrt[4]{x}\cdot\sqrt[8]{x}\cdot\cdot\cdot \sqrt[2^{k}]{x}\)

\(\displaystyle =\prod_{k=0}^{\infty}\sqrt[2^{k}]{x}=\Large {x^{\sum_{k=0}^{\infty}\frac{1}{2^{k}}}}\)

So, we must evaluate \(\displaystyle \sum_{k=0}^{\infty}\frac{1}{2^{k}}\).

What does this evaluate to?. Looks like a basic geometric series.

 

[123]

I don't know this solving

\(\displaystyle =\prod_{k=0}^{\infty}\sqrt[2^{k}]{x}=\Large {x^{\sum_{k=0}^{\infty}\frac{1}{2^{k}}}}\)
...

Isn't other way?

 

[soroban]

Hello, 123!

Galactus is correct.
Don't be intimidated by the notation.

\(\displaystyle \text{Given the equation: }\:x\cdot \sqrt{x}\cdot \sqrt[4]{x}\cdot \sqrt[8]{x}\cdot \sqrt[16]{x}\:\hdots \:=\:256\)

\(\displaystyle \text{(1) Show that: }x\cdot \sqrt{x}\cdot \sqrt[4]{x}\cdot \sqrt[8]{x}\cdot \sqrt[16]{x}\:\hdots \:\:=\:x^2\)

\(\displaystyle \text{We have: }\; x\cdot \sqrt{x}\cdot \sqrt[4]{x}\cdot \sqrt[8]{x}\cdot \sqrt[16]{x}\:\hdots \;=\; x\cdot x^{\frac{1}{2}} \cdot x^{\frac{1}{4}} \cdot x^{\frac{1}{8}}\cdot x^{\frac{1}{16}}\:\cdots \;=\;x^{(1+\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \hdots)}\)

\(\displaystyle \text{That exponent is a geomtric series with first term }a = 1\text{ and common ratio }r = \tfrac{1}{2}\)
. . \(\displaystyle \text{Hence, its sum is: }\:\frac{1}{1-\frac{1}{2}} \,=\,2\)

. . \(\displaystyle \text{Therefore: }\:x\cdot \sqrt{x}\cdot \sqrt[4]{x}\cdot \sqrt[8]{x}\cdot \sqrt[16]{x}\:\hdots \;=\;x^2\)

\(\displaystyle \text{(2) Solve the equation.}\)

\(\displaystyle \text{We have: }\:x^2 \:=\:256\)

\(\displaystyle \text{Therefore: }\:x \,=\,16\)