Question

If ${2}^{ x + 1} = {3}^{ 1 - x}$then find the value of $x.$


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Answer 1

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Given :

• ${2}^{ x + 1} = {3}^{ 1 - x}$

To Find :

• value of x

Solution :

$\sf\huge\rightarrowtail$ ${2}^{ x + 1} = {3}^{ 1 - x}$

Taking $\log$ both sides ,

$\log$ ${2}^{ x + 1} = [tex]\log$ ${3}^{ 1 - x}$

$\sf\huge\implies$ ( 1 + x ) log² = ( 1 - x ) log³

{ $\because$ log $a^{b}$ = b log a }

$\rightarrow$ (1 + x) (0.3010) = (1 - x) (0.477)

{ $\because$ loge = $\color{red} {0.3010}$ =

, log3 = $\color{red} {0.477}$ =

)

$\sf\huge\implies$ $\frac{1 + x}{1 - x}$ = $\color{red} {1.58}$

$\sf\huge\implies$$\frac{1 + x}{1 - x}$ = $\color{red} {1.6}$

$\sf\huge\implies$ 1.6 - 1.6 = 1 + x

$\sf\huge\implies$ $\color{red} {0.6}$ = $\color{red} {2.6x}$

$\sf\huge\implies$ x = $\frac{0.6}{2.6}$

$\sf\huge\implies$ x = $\color{green} {0.23}$

Hence, value of x is 0.23

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Final answer :- value of x = 0.23

Note : swipe left to right to see full answer

Answer 2

_____________________________

Given :

${2}^{ x + 1} = {3}^{ 1 - x}$

To Find :

value of x

Solution :

$\sf\huge\rightarrowtail$ ${2}^{ x + 1} = {3}^{ 1 - x}$

Taking $\log$ both sides ,

$\log$ ${2}^{ x + 1} = [tex]\log$ ${3}^{ 1 - x}$

$\sf\huge\implies$ ( 1 + x ) log² = ( 1 - x ) log³

{ $\because$ log $a^{b}$ = b log a }

$\rightarrow$ (1 + x) (0.3010) = (1 - x) (0.477)

{$\because$ log2 = 0.3010 , log3 = 0.477}

$\sf\huge\implies$ $\frac{1 + x}{1 - x}$ = 1.58

$\sf\huge\implies$ ] $\frac{1 + x}{1 - x}$ = 1.6

$\sf\huge\implies$ 1.6 - 1.6 = 1 + x

$\sf\huge\implies$ 0.6 = 2.6x

$\sf\huge\implies$ x = $\frac{0.6}{2.6}$

$\sf\huge\implies$ x = 0.23

Hence, the value of x = 0.23

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$\color{red} {Final\:Answer}$ = $\color{green} {0.23}$