Question

Y = 10 ^ (( 1/( 1 - log X base 10)) and Z = 10 ^ (( 1/ ( 1- log y base 10 ) then prove that X = 10^[ 1/ ( 1- log Z base 10)]​

Answer 1

$\underline{\text{Solution :}}$

$\mathrm{Given,\:Y=10^{\frac{1}{1-log_{10}X}}}$

$\to \mathrm{log_{10}Y=log_{10}10^{\frac{1}{1-log_{10}X}}}$

$\to \mathrm{log_{10}Y=\frac{1}{1-log_{10}X}\:...(i)}$

$\mathrm{and\:Z=10^{\frac{1}{1-log_{10}Y}}}$

$\to \mathrm{log_{10}Z=log_{10}10^{\frac{1}{1-log_{10}Y}}}$

$\to \mathrm{log_{10}Z=\frac{1}{1-log_{10}Y}}$

$\to \mathrm{log_{10}Z=\frac{1}{1-\frac{1}{1-log_{10}X}}\:by\:(i)}$

$\to \mathrm{1-\frac{1}{1-log_{10}X}=\frac{1}{log_{10}Z}}$

$\to \mathrm{\frac{1}{1-log_{10}X}=1-\frac{1}{log_{10}Z}}$

$\to \mathrm{\frac{1}{1-log_{10}X}=\frac{log_{10}Z-1}{log_{10}Z}}$

$\to \mathrm{1-log_{10}X=\frac{log_{10}Z}{log_{10}Z-1}}$

$\to \mathrm{log_{10}X=1-\frac{log_{10}Z}{log_{10}Z-1}}$

$\to \mathrm{log_{10}X=\frac{log_{10}Z-1-log_{10}Z}{log_{10}Z-1}}$

$\to \mathrm{log_{10}X=\frac{-1}{log_{10}Z-1}}$

$\to \mathrm{log_{10}X=\frac{1}{1-log_{10}Z}}$

$\to \mathrm{10^{log_{10}X}=10^{\frac{1}{1-log_{10}Z}}}$

$\to \boxed{\mathrm{X=10^{\frac{1}{1-log_{10}Z}}}}$

$\text{Hence, proved.}$

Answer 2

Answer:

See photos.

Step-by-step explanation:

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