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Find x for given algebraic expression ${{2}^{2x+2}}-{{6}^{x}}-2\times {{3}^{2x+2}}=0$.

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Last updated date: 27th Apr 2024
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Answer
VerifiedVerified
416.1k+ views
Hint: We will be using the concept of exponents to simplify the problem then we will use the method of forming quadratic equation by substitution to further solve the question.

Complete step-by-step solution -
Now, we have been given that ${{2}^{2x+2}}-{{6}^{x}}-2\times {{3}^{2x+2}}=0$.
Now, we know that ${{2}^{a+b}}={{2}^{a}}\times {{2}^{b}}$. So, we have,
${{2}^{2x}}\times {{2}^{2}}-{{6}^{x}}-2\times {{3}^{2x}}\times {{3}^{2}}=0$
Now, we will divide the whole equation by ${{3}^{2x}}$. So, that we have,
$\dfrac{{{2}^{2x}}\times {{2}^{2}}}{{{3}^{2x}}}-\dfrac{-{{6}^{x}}}{{{3}^{2x}}}-\dfrac{2\times {{3}^{2x}}\times {{3}^{2}}}{{{3}^{2x}}}=0$
Now, on simplifying the equation we have,
$\begin{align}
  & 4{{\left( \dfrac{{{2}^{x}}}{{{3}^{x}}} \right)}^{2}}-\dfrac{{{2}^{x}}\times {{3}^{x}}}{{{3}^{x}}\times {{3}^{x}}}-2\times {{3}^{2}}=0 \\
 & 4{{\left( \dfrac{{{2}^{x}}}{{{3}^{x}}} \right)}^{2}}-\dfrac{{{2}^{x}}}{{{3}^{x}}}-18=0 \\
\end{align}$
Now, we will let $\dfrac{{{2}^{x}}}{{{3}^{x}}}=y$. So, we have,
$4{{y}^{2}}-y-18=0$
Now, we will solve the quadratic equation for the value of y with the help of factorization method. So, we have,
$\begin{align}
  & 4{{y}^{2}}+8y-9y-18=0 \\
 & 4y\left( y+2 \right)-9\left( y+2 \right)=0 \\
 & \left( 4y-9 \right)\left( y+2 \right)=0 \\
 & either\ y=\dfrac{+9}{4}\ or\ y=-2 \\
\end{align}$
Now, we will find the value of x by substituting the value of y as,
$\left( \dfrac{{{2}^{x}}}{{{3}^{x}}} \right)=-2$
So, for \[y=-2\], we don’t have any value of x. Since for any value of $x\in R\ \ \ {{a}^{x}}\ge 0$.
So, now we will solve for $y=\dfrac{9}{4}$.
$\begin{align}
  & \dfrac{{{2}^{x}}}{{{3}^{x}}}=\dfrac{9}{4} \\
 & {{\left( \dfrac{2}{3} \right)}^{x}}={{\left( \dfrac{3}{2} \right)}^{2}} \\
 & \Rightarrow {{\left( \dfrac{2}{3} \right)}^{x}}={{\left( \dfrac{3}{2} \right)}^{-2}} \\
\end{align}$
Since, we know that ${{a}^{-x}}=\dfrac{1}{{{a}^{x}}}$.
So, on comparing we have $x=-2$.

Note: To solve these types of questions it is important to note the way we have made a quadratic equation from the given question by dividing the whole equation with \[{{3}^{2x}}\]. Here we can use the quadratic formula for finding the value of y.