If $x 1,\ y 1,\ \mathcal{z} 1$ are in GP, then $\frac{1}{1 + \log x},\frac{1}{1 + \log y},\frac{1}{1 + \log\mathcal{z}}$ are in

Question

If $x > 1,\ y > 1,\ \mathcal{z} > 1$ are in GP, then $\frac{1}{1 + \log x},\frac{1}{1 + \log y},\frac{1}{1 + \log\mathcal{z}}$ are in

Dr. Abubacker · Accepted Answer

(b)
Since, $x,\ y,\ \mathcal{z}$ are in GP.
$	herefore\ \ y^{2} = x\mathcal{z}$
$\Rightarrow \ 2\log{y = \log{x + \log\mathcal{z}}}$
$\Rightarrow \ \ 2(\log{y + 1) = (1 + \log{x) + (1 + \log\mathcal{z)}}}$
$\Rightarrow \ \ 1 + \log{x,\ 1 + \log{y,\ 1 + \log\mathcal{z}\ }}$ are in AP
$\Rightarrow \ \frac{1}{1 + \log x},\frac{1}{1 + \log y},\frac{1}{1 + \log\mathcal{z}}$ are in HP.

Q. If $x > 1,\ y > 1,\ \mathcal{z} > 1$ are in GP, then $\frac{1}{1 + \log x},\frac{1}{1 + \log y},\frac{1}{1 + \log\mathcal{z}}$ are in

Explanation

Reviewed by Dr. Abubacker

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