Question

solve the following equation for x . give me solution as soon as possible .​

Answer 1

Answer:

${( \sqrt[3]{4}) }^{ \frac{4x + 1}{2} } = {2}^{ - 5}$

$= > {({4}^{ \frac{1}{3} })}^{ \frac{4x + 1}{2} } = {2}^{ - 5}$

$= > {2}^{ \frac{2}{3} \times \frac{4x + 1}{2} } = {2}^{ - 5}$

$= > {2}^{ \frac{4x + 1}{3} } = {2}^{ - 5}$

On equating the powers,

$\frac{4x + 1}{3} = - 5$

$= > 4x + 1 = - 15$

$= > 4x = - 16$

$= > x \frac{ - 16}{4} = - 4$

Hope it helps you.