Question

What is the value of
x in 32^x*16^-2x=4*2^-5x​

Answer 1

GIVEN :–

$\\ \implies{ \bold{ ({32}^{x})( {16}^{ - 2x}) = 4( {2}^{ - 5x})}}$

TO FIND :–

• Value of 'x' = ?

SOLUTION :–

$\\ \implies{ \bold{ ({32}^{x})( {16}^{ - 2x}) = 4( {2}^{ - 5x})}}$

• We know that –

$\\ \to \: \: { \bold{ {2}^{1} = 2}}$

$\\ \to \: \: { \bold{ {2}^{2} = 4}}$

$\\ \to \: \: { \bold{ {2}^{3} = 8}}$

$\\ \to \: \: { \bold{ {2}^{4} = 16}}$

$\\ \to \: \: { \bold{ {2}^{5} = 32}}$

• So , We should write this as –

$\\ \implies{ \bold{ \{({ {2}^{5} )}^{x} \} \{({ {2}^{4}) }^{ - 2x} \} = ( {2}^{2} )( {2}^{ - 5x})}}$

• Using identity –

$\\ \implies{ \bold{ ({a}^{b})^{c} = {a}^{bc} }}$

• So –

$\\ \implies{ \bold{({ {2}^{5x} )} ({ {2}^{ - 8x}) }= ( {2}^{2} )( {2}^{ - 5x})}}$

• Using identity –

$\\ \implies{ \bold{ {a}^{b}.{a}^{c} = {a}^{b + c} }}$

$\\ \implies{ \bold{({ {2}^{5x - 8x} )} = ( {2}^{2 - 5x})}}$

$\\ \implies{ \bold{{ {2}^{ - 3x}} = {2}^{2 - 5x}}}$

• Now compare –

$\\ \implies{ \bold{ - 3x = 2 - 5x}}$

$\\ \implies{ \bold{5x - 3x = 2}}$

$\\ \implies{ \bold{2x = 2}}$

$\\ \implies \large{ \boxed{ \bold{x = 1}}}$

Answer 2

SOLUTION :

${\tt{32^x \:×\: 16^(-2x) \:= 4 \:×\: 2^(-5x)}}$

$\longrightarrow$ ${\tt{2^(5x)\: × \:2^(-8x) \:= \:2^2\: - 2^(-5x)}}$

$\longrightarrow$ ${\tt{2^(-3x) \: =\: 2^(-5x +2)}}$

$\longrightarrow$ ${\tt{ - 3x \:= \:-5x\: +\: 2}}$

$\longrightarrow$ ${\tt{ -3x \:+\: 5x\: = \:2}}$

$\longrightarrow$ ${\tt{ 2x\: = 2 }}$

$\longrightarrow$ ${\tt{\red{ x = 1}}}$

Some exponents formulas,

•a^0=1

•a^1=a

•√a=a^1/2

•(a^m)^p = a^(mp)

•a^ma^n=a^(m+n)