Question

In the equation ${{4}^{x+2}}={{2}^{x+3}}+48$, find the value of $x$ .

Answer 1

Hint: To find the value of $x$ in the equation ${{4}^{x+2}}={{2}^{x+3}}+48$ , we have to write the equation as ${{\left( {{2}^{2}} \right)}^{x+2}}={{2}^{x+3}}+48$ . Using rules of exponents, we will get ${{2}^{2x+4}}={{2}^{x+3}}+48$ . Using ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ , ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ and simplifying, we will get \[2{{\left( {{2}^{x}} \right)}^{2}}-\left( {{2}^{x}} \right)=6\] . Let us consider $y={{2}^{x}}$ . Hence, we will get a polynomial, $2{{x}^{2}}-x-6=0$. Find the roots of this polynomial and so we get $y=2,\dfrac{-3}{2}$ . Substitute the value of y in this and consider \[{{2}^{x}}={{2}^{1}}\] . From this, the value of x can be easily found.

Complete step-by-step solution:
We have to find the value of $x$ in the equation ${{4}^{x+2}}={{2}^{x+3}}+48$ . We will use the rules of exponents here.
We know that 4 can be written as ${{2}^{2}}$ . Hence, the given equation becomes
${{\left( {{2}^{2}} \right)}^{x+2}}={{2}^{x+3}}+48$
We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ .So, the above equation can be written as
${{2}^{2\left( x+2 \right)}}={{2}^{x+3}}+48$
Let us multiply the exponents in the first term. We will get
${{2}^{2x+4}}={{2}^{x+3}}+48$
We know that ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ . Let’s apply this in LHS and RHS. We will get
\[{{2}^{4}}\times {{2}^{2x}}=\left( {{2}^{3}}\times {{2}^{x}} \right)+48\]
We know that ${{2}^{4}}=2\times 2\times 2\times 2=16$ and ${{2}^{3}}=2\times 2\times 2=8$ . Hence, the above equation becomes
\[16\times {{2}^{2x}}=\left( 8\times {{2}^{x}} \right)+48\]
Let us rearrange the terms. We will get
\[\left( 16\times {{2}^{2x}} \right)-\left( 8\times {{2}^{x}} \right)=48\]
We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ . Hence, we can write the previous equation as
\[16{{\left( {{2}^{x}} \right)}^{2}}-8\left( {{2}^{x}} \right)=48\]
Let us take 8 common from LHS. We will get
\[8\left[ 2{{\left( {{2}^{x}} \right)}^{2}}-\left( {{2}^{x}} \right) \right]=48\]
Now, we can take 8 to the RHS.
\[\begin{align}
  & \Rightarrow 2{{\left( {{2}^{x}} \right)}^{2}}-\left( {{2}^{x}} \right)=\dfrac{48}{8} \\
 & \Rightarrow 2{{\left( {{2}^{x}} \right)}^{2}}-\left( {{2}^{x}} \right)=6 \\
\end{align}\]
Let’s take 6 to RHS.
\[\Rightarrow 2{{\left( {{2}^{x}} \right)}^{2}}-\left( {{2}^{x}} \right)-6=0\]
Let us consider $y={{2}^{x}}$ . Hence, we will get a polynomial,
$2{{x}^{2}}-x-6=0$
Let’s solve this equation.
We know that $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for a polynomial $a{{x}^{2}}+bx+c=0$ . Hence,
$y=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-\left( 4\times 2\times -6 \right)}}{2\times 2}$
Let’s solve this. We will get
$\begin{align}
  & y=\dfrac{1\pm \sqrt{1+48}}{4} \\
 & \Rightarrow y=\dfrac{1\pm \sqrt{49}}{4} \\
\end{align}$
By taking the root, we will get
$y=\dfrac{1\pm 7}{4}$
This means that
$\begin{align}
  & y=\dfrac{1+7}{4},\dfrac{1-7}{4} \\
 & \Rightarrow y=\dfrac{8}{4},\dfrac{-6}{4} \\
 & \Rightarrow y=2,\dfrac{-3}{2} \\
\end{align}$
Now we can substitute the value of y. We will get
${{2}^{x}}=2,\dfrac{-3}{2}$
Let us consider ${{2}^{x}}=2$
We can write this as
\[{{2}^{x}}={{2}^{1}}\]
We know that when bases are equal, their powers will also be equal. Hence,
\[x=1\]
Hence, the value of $x$ is 1.

Note: You must know the rules of exponents to solve this problem. You may write the formula for ${{\left( {{a}^{m}} \right)}^{n}}\text{as }{{a}^{m+n}}$ , ${{a}^{m}}\times {{a}^{n}}\text{ as }{{a}^{m-n}}$ thus making error in the solution. From the step \[16{{\left( {{2}^{x}} \right)}^{2}}-8\left( {{2}^{x}} \right)=48\] , we can solve this problem in an alternate way.
\[\Rightarrow 16{{\left( {{2}^{x}} \right)}^{2}}-8\left( {{2}^{x}} \right)-48=0\]
We can write \[-8\left( {{2}^{x}} \right)\] as $-32\left( {{2}^{x}} \right)+24\left( {{2}^{x}} \right)$ . We will get
\[\Rightarrow 16{{\left( {{2}^{x}} \right)}^{2}}-32\left( {{2}^{x}} \right)+24\left( {{2}^{x}} \right)-48=0\]
Let us take the common terms outside. We will get
\[16\left( {{2}^{x}} \right)\left( {{2}^{x}}-2 \right)+24\left( {{2}^{x}}-2 \right)=0\]
Now, we can take ${{2}^{x}}-2$ outside. We will get
 \[\left( {{2}^{x}}-2 \right)\left( 16\left( {{2}^{x}} \right)+24 \right)=0\]
This mean that, either \[\left( {{2}^{x}}-2 \right)=0\text{ or }\left( 16\left( {{2}^{x}} \right)+24 \right)=0\]
Let us consider \[\left( {{2}^{x}}-2 \right)=0\]
We can rearrange the terms to get
\[{{2}^{x}}=2\]
We know that when bases are equal, their powers will also be equal. Hence,
\[x=1\]