Question

If $2^x=4^y=8^z$ and $\frac{1}{2x}+\frac{1}{4y}+\frac{1}{4z}=4$, find the value of $x$.

Answer 1

Step 1: Find the value of $y\text{and}z$ in terms of $x$.

It is given that, $2^x=4^y=8^z$

Express each side of the equation as a power of the same base and then equate the exponents in order to find the value of $y\text{and}z$ in terms of $x$.

Therefore, using the law of exponent, we get,

$\begin{array}{l}{2}^x=4^y\\ \Rightarrow 2^x=2^{2y}\\ \Rightarrow x=2y\\ \Rightarrow y=\frac{x}{2}\end{array}$

And,

$\begin{array}{l}{2}^x=8^z\\ \Rightarrow 2^x=2^{3z}\\ \Rightarrow x=3z\\ \Rightarrow z=\frac{x}{3}\end{array}$

Step 2: Find the value of $x$.

Substitue the value of $y\text{and}z$ on $\frac{1}{2x}+\frac{1}{4y}+\frac{1}{4z}=4$,

$\begin{array}{l}\frac{1}{2x}+\frac{1}{4y}+\frac{1}{4z}=4\\ \frac{1}{2x}+\frac{1}{4\left(\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}+\frac{1}{4\left(\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}=4\\ \frac{1}{2x}+\frac{1}{2x}+\frac{3}{4x}=4\\ \frac{2+2+3}{4x}=4\\ \frac{7}{4x}=4\\ 16x=7\\ x=\frac{7}{16}\end{array}$

Hence, the value of $x$ is $\frac{7}{16}$.