Question

(-2/3)^-3*(-2/3)^-6=(-2/3)^3x​

Answer 1

Answer:

x=-3

Given:

$\texttt{ (\dfrac{-2}{3})^{-3}\times(\dfrac{-2}{3})^{-6}=(\dfrac{-2}{3})^{3x} }$

To Find:

$\textbf{Value\:of\:x}$

Explanation:

$\textsf{ (\dfrac{-2}{3})^{-3}\times(\dfrac{-2}{3})^{-6}=(\dfrac{-2}{3})^{3x} }$

$\texttt{We know that if the base is same}$$\texttt{the power is added. }$

$\textsf{ (\dfrac{-2}{3})^{(-3+(-6))}=(\dfrac{-2}{3})^{3x} }$

$\textsf{ (\dfrac{-2}{3})^{-9}=(\dfrac{-2}{3})^{3x} }$

$\textsf{Comparing the powers on both sides}$

$\textsf{3x=-9}$

$\textsf{x=-3}$

Additional Information:

$\large\odot\:\:\textbf{If\:we\:have \:to\:multiply \:same}$

$\large\textbf{bases with different power}$

$\large\textbf{,we\:add\:the\:power\:keeping}$

$\textbf{the\:bases\:same}$

$\Large\textbf{\underline{Example}}$

$\textbf{Let\:us\:consider\: a^b\:\times\:a^c }$

$\textbf{where\:a,\:b\:and\:c\:are \:constants.}$

$\textbf{Here\:we\:will\:add\:the\:powers \:b\:and\:c}$

$\textbf{keeping\:the\:base\:a\:same.}$

$\large\textsf{ a^{(b+c)} }$