Solve the following equations for $x$:$ 2^{x+1}=4^{x-3} $

Given:

\( 2^{x+1}=4^{x-3} \)

To do:

We have to find the value of $x$.

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$  

Therefore,

$2^{x+1}=4^{x-3}$

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

$2^{x+1}=2^{2 x-6}$

Comparing the powers on both sides, we get,

$x+1=2 x-6$

$2 x-x=1+6$

$x=7$

Therefore, the value of $x$ is $7$. 

Tutorialspoint

Updated on: 10-Oct-2022

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