Question

If ${{4}^{2x-1}}-{{16}^{x-1}}=384$, find the value of $x$. \[\]

Answer 1

Hint: e covert 16 in the left hand side as a power of 4. We use law of exponents ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}={{a}^{mn}}$ to proceed and then take ${{4}^{2x-1}}$ common. We simplify until we get same base in left and right side of the equation, then equate the exponents of both side and then solve for $x$. \[\]

Complete step by step answer:
We know from the law of exponents that when the base is the same power raised to another power we can keep the same base and multiply the power. If $a$ is the base is raised to a power $m$as ${{a}^{m}}$and ${{a}^{m}}$is raised to a power $n$as ${{\left( {{a}^{m}} \right)}^{n}}$ we can write as
\[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}={{a}^{mn}}\]
We know from law of exponents of division with same base that that when we are dividing two numbers with same base then we can write with same base and power of divisor subtracted from power of dividend. It means
\[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\]
We are given the equation where unknown $x$ is in the exponents as
\[{{4}^{2x-1}}-{{16}^{x-1}}=384\]
We see that we can only solve for $x$ when the bases at the left hand side and right will be same. So let us first convert 16 in the left hand side in the form of power of 4. We have;
\[\begin{align}
  & \Rightarrow {{4}^{2x-1}}-{{\left( 4\times 4 \right)}^{x-1}}=384 \\
 & \Rightarrow {{4}^{2x-1}}-{{\left( {{4}^{2}} \right)}^{x-1}}=384 \\
\end{align}\]
We use the law of exponents ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}={{a}^{mn}}$ for $a=2,m=4,n=x$ in the left hand side to have;
\[\begin{align}
  & \Rightarrow {{4}^{2x-1}}-{{4}^{2\left( x-1 \right)}}=384 \\
 & \Rightarrow {{4}^{2x-1}}-{{4}^{2x-2}}=384 \\
 & \Rightarrow {{4}^{2x-1}}-{{4}^{\left( 2x-1 \right)-1}}=384 \\
\end{align}\]
We use the law of exponents $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ for $a=4,m=2x-1,n=1$ in the left hand side second term to have;
\[\begin{align}
  & \Rightarrow {{4}^{2x-1}}-\dfrac{{{4}^{2x-1}}}{{{4}^{1}}}=384 \\
 & \Rightarrow {{4}^{2x-1}}-\dfrac{{{4}^{2x-1}}}{4}=384 \\
\end{align}\]
We take ${{4}^{2x-1}}$ common in the left hand side of the above step to have;
\[\begin{align}
  & \Rightarrow {{4}^{2x-1}}\left( 1-\dfrac{1}{4} \right)=384 \\
 & \Rightarrow {{4}^{2x-1}}\left( \dfrac{4-1}{4} \right)=384 \\
 & \Rightarrow {{4}^{2x-1}}\times \dfrac{3}{4}=384 \\
\end{align}\]
We multiply $\dfrac{4}{3}$ in both side of the above step to have;
\[\begin{align}
  & \Rightarrow {{4}^{2x-1}}\times \dfrac{3}{4}\times \dfrac{4}{3}=384\times \dfrac{4}{3} \\
 & \Rightarrow {{4}^{2x-1}}=512 \\
\end{align}\]
Let us express 4 and 512 as powers of 2. We have;
\[\Rightarrow {{\left( {{2}^{2}} \right)}^{2x-1}}={{2}^{9}}\]
We use the law of exponents ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}={{a}^{mn}}$ for $a=2,m=2,n=2x-1$ in the left hand side to have;
\[\begin{align}
  & \Rightarrow {{2}^{2\times \left( 2x-1 \right)}}={{2}^{9}} \\
 & \Rightarrow {{2}^{4x-2}}={{2}^{9}} \\
\end{align}\]
We equate the exponents of both sides to have;
\[\begin{align}
  & \Rightarrow 4x-2=9 \\
 & \Rightarrow 4x=11 \\
 & \Rightarrow x=\dfrac{11}{4} \\
\end{align}\]

So the value of $x$ is found to be $\dfrac{11}{4}$.

Note: We note that when we needed the same base both left and right hand side of \[{{4}^{2x-1}}=512\] but we have $512={{2}^{9}}$ which cannot be expressed as a power of 4 that is why we expressed both the terms in the power of 2. We can also directly convert 4 and 16 in the power of 2 and the use law of exponents $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$. We must be careful that base and power simultaneously cannot be zero.