Question

Find the value of x , where x is an integer $\dfrac{{{2}^{2x}}\times 4\times {{2}^{x}}-{{8}^{x}}}{{{\left( {{2}^{5}} \right)}^{3}}\times 9}=\dfrac{1}{24}$ .

Answer 1

Hint: To find the value of x in $\dfrac{{{2}^{2x}}\times 4\times {{2}^{x}}-{{8}^{x}}}{{{\left( {{2}^{5}} \right)}^{3}}\times 9}=\dfrac{1}{24}$ , we must simplify the expression using laws of exponents. Using ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ , we will get $\dfrac{{{2}^{3x}}\times 4-{{\left( {{2}^{3}} \right)}^{x}}}{{{\left( {{2}^{5}} \right)}^{3}}\times 9}=\dfrac{1}{24}$ and using ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ , we can further simplify the expression to $\dfrac{{{2}^{3x}}\times 4-{{2}^{3x}}}{{{\left( {{2}^{5}} \right)}^{3}}\times 9}=\dfrac{1}{24}$ . This can be written as $\dfrac{{{2}^{3x}}\times 3}{{{\left( {{2}^{5}} \right)}^{3}}\times 9}=\dfrac{1}{24}$ . By solving this, we will get ${{2}^{3x}}=\dfrac{{{\left( {{2}^{5}} \right)}^{3}}\times 3}{24}=\dfrac{{{\left( {{2}^{5}} \right)}^{3}}}{{{2}^{3}}}$ . Use the formula $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ to get ${{2}^{3x}}={{2}^{12}}$ . We know that if the bases are equal on both the sides, then their powers are equal. Hence, $3x=12$ . Solving this gives the value of x.

Complete step by step answer:
We have to find the value of x in $\dfrac{{{2}^{2x}}\times 4\times {{2}^{x}}-{{8}^{x}}}{{{\left( {{2}^{5}} \right)}^{3}}\times 9}=\dfrac{1}{24}$ . Let us simplify this equation.
We know that ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ . Hence, we can write the above equation as
$\dfrac{{{2}^{3x}}\times 4-{{8}^{x}}}{{{\left( {{2}^{5}} \right)}^{3}}\times 9}=\dfrac{1}{24}$
Now, we can write 8 in terms of power as ${{2}^{3}}=2\times 2\,\times 2=8$ . Hence,
$\dfrac{{{2}^{3x}}\times 4-{{\left( {{2}^{3}} \right)}^{x}}}{{{\left( {{2}^{5}} \right)}^{3}}\times 9}=\dfrac{1}{24}$
We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ . Hence, the above equation can be written as
$\dfrac{{{2}^{3x}}\times 4-{{2}^{3x}}}{{{\left( {{2}^{5}} \right)}^{3}}\times 9}=\dfrac{1}{24}$
Now, let us take ${{2}^{3x}}$ outside the numerator. We will get
$\dfrac{{{2}^{3x}}\left( 4-1 \right)}{{{\left( {{2}^{5}} \right)}^{3}}\times 9}=\dfrac{1}{24}$
We can write this as
$\dfrac{{{2}^{3x}}\times 3}{{{\left( {{2}^{5}} \right)}^{3}}\times 9}=\dfrac{1}{24}$
Let us cancel the common factors from numerator and denominator. We will get
$\dfrac{{{2}^{3x}}}{{{\left( {{2}^{5}} \right)}^{3}}\times 3}=\dfrac{1}{24}$
Let us take the denominator to the RHS. We will get
${{2}^{3x}}=\dfrac{{{\left( {{2}^{5}} \right)}^{3}}\times 3}{24}$
We can now solve the RHS by cancelling the common factors.
${{2}^{3x}}=\dfrac{{{\left( {{2}^{5}} \right)}^{3}}}{8}$
Now, let us write 8 in terms of powers of 2. We will get
${{2}^{3x}}=\dfrac{{{\left( {{2}^{5}} \right)}^{3}}}{{{2}^{3}}}$
We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ . Hence, we can write the above expression as
${{2}^{3x}}=\dfrac{{{2}^{15}}}{{{2}^{3}}}$
We have studied that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ . So the above expression becomes
$\begin{align}
  & {{2}^{3x}}={{2}^{15-3}} \\
 & \Rightarrow {{2}^{3x}}={{2}^{12}} \\
\end{align}$
We know that if the bases are equal on both the sides, then their powers are equal. Hence,
$\begin{align}
  & 3x=12 \\
 & \Rightarrow x=\dfrac{12}{3} \\
\end{align}$
By solving this, we will get
$x=4$

Hence the value of x is 4.

Note: You must know the laws of exponents to solve this. Try to make the terms simpler by writing it in terms of exponents. You may make mistake when writing the formula ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ as ${{a}^{m}}\times {{a}^{n}}={{a}^{m-n}}$ . Similarly, you may write $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ as $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m+n}}$ .