Question

If $a^x=b^y=c^z$ and $b^2=ac$ then $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$

• A. True
• B. False

Answer 1

The correct option is B

False

Given $a^x=b^y=c^z$ and $b^2=ac$

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

$\therefore a=k^{\frac{1}{x}},b=k^{\frac{1}{y}},c=k^{\frac{1}{z}}$

Given $b^2=ac$

$\Rightarrow k^{\frac{2}{y}}=k^{\frac{1}{x}}\times k^{\frac{1}{z}}$

$\Rightarrow k^{\frac{2}{y}}=k^{\frac{1}{x}+\frac{1}{z}}$ (product law)

Since bases are equal, exponents may be equal

$\frac{2}{y}=\frac{1}{x}+\frac{1}{z}$

Hence the given statement is false