Question

Solve the following equation for x, $9\times {{81}^{x}}={{\left( 27 \right)}^{2-x}}$.

Answer 1

Hint: We will first replace 9 by ${{3}^{2}}$, 81 by ${{3}^{4}}$ and 27 by ${{3}^{3}}$. Now we will use ${{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}},{{a}^{x}}.{{a}^{y}}={{a}^{x+y}},\dfrac{{{a}^{x}}}{{{a}^{y}}}={{a}^{x-y}}$ to simplify the equation and finally write 1 as ${{\left( 3 \right)}^{0}}$ and equate the powers to get the value of x

Complete step-by-step answer:
Here, we have to find the value of x by solving the equation $9\times {{81}^{x}}={{\left( 27 \right)}^{2-x}}$. So, to solve this, let us consider the equation given in the question,
$9\times {{81}^{x}}={{\left( 27 \right)}^{2-x}}\ldots \ldots \ldots \left( i \right)$
Now, we know that we can write 9, 27 and 81 as follows,
$\begin{align}
  & 3\times 3=9={{3}^{2}} \\
 & 3\times 3\times 3=27={{3}^{3}} \\
 & 3\times 3\times 3\times 3=81={{3}^{4}} \\
\end{align}$
So, we will replace 9 by ${{3}^{2}}$, 81 by ${{3}^{4}}$ and 27 by ${{3}^{3}}$in equation (i). So, we will get the equation as,
$\left( {{3}^{2}} \right)\times {{\left( {{3}^{4}} \right)}^{x}}={{\left( {{3}^{3}} \right)}^{2-x}}$
We know that ${{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}}$. So, we will apply it in the above equation. So, we get,
$\begin{align}
  & \left( {{3}^{2}} \right)\times {{\left( 3 \right)}^{4x}}={{\left( 3 \right)}^{3\left( 2-x \right)}} \\
 & \Rightarrow \left( {{3}^{2}} \right)\times {{\left( 3 \right)}^{4x}}={{\left( 3 \right)}^{6-3x}} \\
\end{align}$
Now, we know that ${{a}^{x}}.{{a}^{y}}={{a}^{x+y}}$. So, we will use this identity in the above equation. So, we get,
${{\left( 3 \right)}^{2+4x}}={{\left( 3 \right)}^{6-3x}}$
We know that when a = b, then $\dfrac{a}{b}=1$. So, by using this relation in the above equation, we will get,
$\dfrac{{{\left( 3 \right)}^{2+4x}}}{{{\left( 3 \right)}^{6-3x}}}=1$
Now, we also know that, $\dfrac{{{a}^{x}}}{{{a}^{y}}}={{a}^{x-y}}$. So, we will use it in the above equation. And we get,
$\begin{align}
  & {{\left( 3 \right)}^{\left( 2+4x \right)-\left( 6-3x \right)}}=1 \\
 & \Rightarrow {{\left( 3 \right)}^{\left[ 2-6+4x+3x \right]}}=1 \\
 & \Rightarrow {{\left( 3 \right)}^{\left[ -4+7x \right]}}=1 \\
\end{align}$
We can write 1 as ${{\left( 3 \right)}^{0}}$. So, we will write it in the above equation and get,
${{\left( 3 \right)}^{\left[ -4+7x \right]}}={{\left( 3 \right)}^{0}}$
When ${{a}^{x}}={{a}^{y}}$, then x = y. By using this relation, we get,
$\begin{align}
  & -4+7x=0 \\
 & \Rightarrow 7x=4\Rightarrow x=\dfrac{4}{7} \\
\end{align}$
Hence, we get the value of x as $\dfrac{4}{7}$.

Note: In this question, the students can verify their answer by substituting the values of x in the given equation and checking if the LHS = RHS as follows:
$\begin{align}
  & 9\times {{\left( 81 \right)}^{\dfrac{4}{7}}}={{\left( 27 \right)}^{2-\dfrac{4}{7}}} \\
 & \Rightarrow {{\left( 3 \right)}^{2}}\times {{\left( 3 \right)}^{4\times \dfrac{4}{7}}}={{\left( 3 \right)}^{3\times \left( \dfrac{10}{7} \right)}} \\
 & \Rightarrow {{\left( 3 \right)}^{2+\dfrac{16}{7}}}={{\left( 3 \right)}^{\dfrac{30}{7}}} \\
 & \Rightarrow {{\left( 3 \right)}^{\dfrac{30}{7}}}={{\left( 3 \right)}^{\dfrac{30}{7}}} \\
\end{align}$
So, since the LHS = RHS, our answer is correct.