Question

If 2's power x =3's power y=36's z then find the value of z

Answer 1

so,is it the right answer?

Answer 2

Given :

The equation as  $2^{x}$ =  $3^{y}$  = $36^{z}$

To Find :

The value of z

Solution :

Let   $2^{x}$ =  $3^{y}$  = $36^{z}$   = k

i.e    $2^{x}$  = k

Or,   $( 2^{x} ) ^{\dfrac{1}{x} }$  = $( k) ^{\dfrac{1}{x} }$

Or,   $( k) ^{\dfrac{1}{x} }$ = 2                  ...............1

Again

i.e    $3^{y}$  = k

Or,   $( 3^{y} ) ^{\dfrac{1}{y} }$  = $( k) ^{\dfrac{1}{y} }$

Or,  $( k) ^{\dfrac{1}{y} }$  = 3                   ...............2

Again

i.e    $36^{z}$  = k

Or,   $( 36^{z} ) ^{\dfrac{1}{z} }$  = $( k) ^{\dfrac{1}{z} }$

Or,  $( k) ^{\dfrac{1}{z} }$  = 36                   ...............3

Now,

From eq 3

$( k) ^{\dfrac{1}{z} }$  =  6 × 6

Or,  $( k) ^{\dfrac{1}{z} }$  =  2² × 3²

Fro, eq 1 and eq 2

$( k) ^{\dfrac{1}{z} }$  =  ( $( k) ^{\dfrac{1}{x} }$  )²  ×  ( $( k) ^{\dfrac{1}{y} }$ ) ²

Or,  $( k) ^{\dfrac{1}{z} }$  =   $( k) ^{\dfrac{2}{x} }$    ×   $( k) ^{\dfrac{2}{y} }$

Or,  $( k) ^{\dfrac{1}{z} }$  =   $k [^{\dfrac{2}{x} +\dfrac{2}{y} }]$               ( ∵ $a^{c}$ × $a^{b}$ = $a^{c+b}$  , from base indices )

As both side have k as base

∴    $\dfrac{1}{z } = \dfrac{2}{x} + \dfrac{2}{y}$

Or,  $\dfrac{1}{z}$  = $\dfrac{2y+2x}{yx}$

Or,  z = $\dfrac{xy}{2(x+y)}$

Hence, The value of z for the given equation is $\dfrac{xy}{2(x+y)}$  Answer