Question

If log 2 ^ 2 (x+1)+log 2 ^ 2 (log 3 y)+log 2 ^ 2 (1+log 2 z)=0 , then the value of (x + y + z) is​

Answer 1

Answer:

4

Step-by-step explanation:

Given ,

$\left(\log_{2}{(x+1)}\right)^2+\left(\log_{2}{\left(\log_{3}{y}\right)}\right)^2+\left(\log_{2}{\left(1+\log_{2}{z}\right)}\right)^2=0$

To Find :-

Value of (x + y + z)

How To Do :-

As they given that

$\left(\log_{2}{(x+1)}\right)^2+\left(\log_{2}{\left(\log_{3}{y}\right)}\right)^2+\left(\log_{2}{\left(1+\log_{2}{z}\right)}\right)^2=0$

if and only if all the terms become '0'. So we need to equate each and every term to '0' and by equating those terms to 'o' we will get the values of 'x , y ' z' separately so we need to add those values to get 'x + y + z'.

Solution :-

$\left(\log_{2}{(x+1)}\right)^2+\left(\log_{2}{\left(\log_{3}{y}\right)}\right)^2+\left(\log_{2}{\left(1+\log_{2}{z}\right)}\right)^2=0$

Equating all terms to '0' :-

$\left(\log_{2}{(x+1)}\right)^2=0,\left(\log_{2}{\left(\log_{3}{y}\right)}\right)^2=0,\left(\log_{2}{\left(1+\log_{2}{z}\right)}\right)^2=0$

First equating the first term :-

$\left(\log_{2}{(x+1)}\right)^2=0$

$\log_{2}({x+1)}=\sqrt{0}$

$\log_{2}{(x+1)}=0$

x + 1 = 2⁰

x + 1 = 1

x = 1 - 1

x = 0

∴ Value of 'x' = 0.

Equating the second term :-

$\left(\log_{2}{\left(\log_{3}{y}\right)}\right)^2=0$

$\log_{2}{\log_{3}{y}}=\sqrt{0}$

$\log_{2}{\log_{3}{y}}=0$

$\log_{3}{y}=2^0$

$\log_{3}{y}=1$

y = 3¹

y = 3

∴ Value of 'y' = 3

Equating the third term to '0' :-

$\left(\log_{2}{\left(1+\log_{2}{z}\right)}\right)^2=0$

$\left(\log_{2}{\left(1+\log_{2}{z}\right)}\right)=\sqrt0$

$\left(\log_{2}{\left(1+\log_{2}{z}\right)}\right)=0$

$1+\log_{2}{z}=2^0$

$1+\log_{2}{z}=1$

$\log_{2}{z}=1-1$

$\log_{2}{z}=0$

z = 2⁰

z = 1

∴ Value of 'z' = 1

x + y + z = 0 + 3 + 1

= 4

∴ x + y + z = 4