If x, y, z are in geometric progression and a | vedclass

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(C) Since $x, y, z$ are in geometric progression,we have $y^2 = xz$. Let $a^x = b^y = c^z = \lambda$. Taking log on both sides,we get $x \log a = y \l

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$log_b\ a = log_c\ b$

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(C) Since $x, y, z$ are in geometric progression,we have $y^2 = xz$. Let $a^x = b^y = c^z = \lambda$. Taking log on both sides,we get $x \log a = y \log b = z \log c = \log \lambda$. Thus,$x = \frac{\log \lambda}{\log a}$,$y = \frac{\log \lambda}{\log b}$,and $z = \frac{\log \lambda}{\log c}$. Substituting these values into $y^2 = xz$,we get: $(\frac{\log \lambda}{\log b})^2 = (\frac{\log \lambda}{\log a}) 	imes (\frac{\log \lambda}{\log c})$. This simplifies to $(\log b)^2 = \log a \cdot \log c$. Dividing both sides by $\log a \cdot \log b$,we get $\frac{\log b}{\log a} = \frac{\log c}{\log b}$,which implies $\log_a b = \log_b c$. Taking the reciprocal,we get $\frac{1}{\log_a b} = \frac{1}{\log_b c}$,which is $\log_b a = \log_c b$.

If $x, y, z$ are in geometric progression and $a^x = b^y = c^z$,then . . . . . .

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