Question

I hate spammers.... Answer this plss (@_@)​

Answer 1

Step-by-step explanation:

Let 2^x = 3^y = 12^z = k.

2^x = k & 3^y = k & 12^z = k

2 = k^(1/x) & 3 = k^(1/y) & 12 = k^(1/z)

As we know, 12 = 2 * 2* 3

=> 12 = 2² * 3

=> k^(1/z) = {k^(1/x)}² * k^(1/y)

=> k^(1/z) = k^(2/x) * k^(1/y)

=> k^(1/z) = k^(2/x + 1/y)

=> 1/z = 2/x + 1/y

=> 1/z - 1/y = 2/x

Answer 2

Step-by-step explanation:

$\large \: \sf \underline{Given \: } :$

• $\sf \: {2^{x}=3^{y}=12^{z}=k}$

$\large \: \sf \underline{To \: Find} :$

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

$\large \: \sf \underline{Solution} :$

$\sf \underline{Let \: us \: consider } \: \: 2^{x}=3^{y}=12^{z}$

Take logarithm on both sides.

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

$</p><p> \sf : \implies \: x=\frac{\log k}{\log 2} ; y=\frac{\log k}{\log 3} ; z=\frac{\log k}{\log 12} \:$

$\sf : \implies\frac{1}{y}+\frac{2}{x}=\frac{\log 3}{\log k}+\frac{2 \log 2}{\log k} \\ \\ \sf : \implies\frac{1}{y}+\frac{2}{x}=\frac{(\log 3+2 \log 2)}{\log k} \: \\ \\ </p><p>$

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

$\sf : \implies \: \: { \frac { (\log 3+\log 4) }{ \log k }}$

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

$\sf : \implies \: \frac{\log 12}{\log k}$

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