Find $ x $ if

$ \left(\frac{1}{3}\right)^{-4} \times\left(\frac{1}{3}\right)^{-8}=\left(\frac{1}{3}\right)^{-4 x} $.

Given:

$\left(\frac{1}{3}\right)^{-4} \times\left(\frac{1}{3}\right)^{-8}=\left(\frac{1}{3}\right)^{-4 x}$.

To do:

We have to find the value of x.

Solution:

We know that,

a^m  x  aⁿ  = a^m+n  

LHS

$\left(\frac{1}{3}\right)^{-4} \times \left(\frac{1}{3}\right)^{-8} =\left(\frac{1}{3}\right)^{-4+( -8)}$

$=\left(\frac{1}{3}\right)^{-( 4+8)}$              [$( +) \times ( -) =( -)$]

$=\left(\frac{1}{3}\right)^{-12}$

RHS = $\left(\frac{1}{3}\right)^{-4 x}$

On comparing LHS and RHS,

$-12=-4x$

$x=\frac{12}{4}$

$x=3$

The value of x is 3.

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Updated on: 10-Oct-2022

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