Question

MATH ARYABHATTAS!
If
${2}^{x} = {3}^{y} = {6}^{z}$
then,
$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} =$
?
​

Answer 1

Answer:

Step-by-step explanation:

Formulas used while solving the question-:

$log x^{y} = ylogx$

$log(x*y) = logx+logy$

Answer is in the image attached. Hope this will help you.!!

Answer 2

Question:-

${2}^{x} = {3}^{y} = {6}^{z} \\ </p><p>then, \\ </p><p>\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = </p><p>$

Answer:-

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{ {log}^{2} }{ {log}^{k} } + \frac{ {log}^{3} }{ {log}^{e} } + \frac{ {log}^{6} }{ {log}^{k} }$

Then,

$= \frac{ {log}^{2} }{ {log}^{k} } + \frac{ {log}^{3} }{ {log}^{k} } + \frac{ {log}^{2} + {log}^{3} }{ {log}^{k} }$

Process,

$= \frac{2}{ {log}^{k} } ( {log}^{2} + {log}^{3)}$

$= 2 \frac{( {log}^{2} + {log}^{3} \: ) }{ {log}^{k} } = \frac{1}{2}$

Answer,

$= \frac{2}{3}$