Question

If 2x=3y=12z,then show that 1/z=1/y+2/x

Answer 1

Solution:

Given

$2^{x}= 3^{y}= (12)^{z}$

$Let \: 2^{x}= 3^{y}= (12)^{z}=k$

i) $2^{x}=k \implies 2=k^{\frac{1}{x}}$--(1)

ii)$3^{y}=k \implies 3=k^{\frac{1}{y}}$---(2)

iii)$(12)^{z}=k \implies 12=k^{\frac{1}{z}}$--(3)

Now ,

$\implies12=k^{\frac{1}{z}}$

$\implies 2^{2}\times 3=k^{\frac{1}{z}}$

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

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

$\boxed { a^{m} × a^{n} = a^{m+n}}$

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

$\boxed {If \: a^{m} = a^{n} \implies m = n}$

Hence , proved

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Answer 2

When $\bold{2^{x}=3^{y}=12^{z}=k}$,  $\bold{\frac{1}{y}+\frac{2}{x}=\frac{1}{z}}$ is proved

Solution:

Let us consider $2^{x}=3^{y}=12^{z}=k$

Take logarithm on both sides.

x log 2 = log k  ; y log 3 = log k ; z log 12 = log k

$x=\frac{\log k}{\log 2} ; y=\frac{\log k}{\log 3} ; z=\frac{\log k}{\log 12}$

$\frac{1}{y}+\frac{2}{x}=\frac{\log 3}{\log k}+\frac{2 \log 2}{\log k}=\frac{(\log 3+2 \log 2)}{\log k}$

${ =\frac { \log 3+\log 2\times 2 }{ \log k } ({ since\ }a\ log\ m=ma) }$

${ =\frac { (\log 3+\log 4) }{ \log k }}$

${ =\frac { \log (3\times 4) }{ \log k } ({ since\ }\log a+\log b=\log ab) }$

$=\frac{\log 12}{\log k}$

Therefore,$\bold{\frac{1}{y}+\frac{2}{x}=\frac{1}{z}}$  is proved.