Question

If $x^a=y^b=z^c$ and $y^2=xz$, then $b=\frac{2ac}{a+c}$

• A. True
• B. False

Answer 1

The correct option is A True

Given: $x^a=y^b=z^c$

Let $x^a=y^b=z^c=k$

$\Rightarrow x=k^{\frac{1}{a}}$

$\Rightarrow y=k^{\frac{1}{b}}$

$\Rightarrow z=k^{\frac{1}{c}}$

Also, given $y^2=xz$

Thus,

$(k^{\frac{1}{b}}{)}^2=k^{\frac{1}{a}}{k}^{\frac{1}{c}}$

$\Rightarrow k^{\frac{2}{b}}=k^{\frac{1}{a}+\frac{1}{c}}$

$\Rightarrow \frac{2}{b}=\frac{1}{a}+\frac{1}{c}$

$\Rightarrow \frac{2}{b}=\frac{a+c}{ac}$

$\Rightarrow b=\frac{2ac}{a+c}$