Question

Solve the given expression $\dfrac{16{{\left( 32 \right)}^{x}}-{{2}^{3x-2}}\times {{4}^{x+1}}}{15\times {{2}^{x-1}}\times {{16}^{x}}}-\dfrac{5\cdot {{5}^{x-1}}}{\sqrt{{{5}^{2x}}}}=$ \[\]
A.1\[\]
B.$-1$\[\]
C.0\[\]
D.4\[\]

Answer 1

Hint: We convert all the composite numbers 16 and 32 into their prime factorization and write them as power of 2. We use the laws of exponents of product with same base ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ , law of exponents involving power raised to the another power${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ and laws of exponents involving quotient $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ to simplify both the terms in the given expression. We simplify until we get an expression of the numerator of the first term where we can take ${{2}^{5x}}$ common and cancel out terms from numerator and denominator.\[\]

Complete step-by-step solution
We are given in the question an expression in $x$ where $x$ occurs in exponents. We have
\[\dfrac{16{{\left( 32 \right)}^{x}}-{{2}^{3x-2}}\times {{4}^{x+1}}}{15\times {{2}^{x-1}}\times {{16}^{x}}}-\dfrac{5\cdot {{5}^{x-1}}}{\sqrt{{{5}^{2x}}}}\]
Let us express all the composite numbers in the expression in their prime factorization. We have,
\[\begin{align}
  & \Rightarrow \dfrac{\left( 2\times 2\times 2\times 2 \right){{\left( 2\times 2\times 2\times 2\times 2 \right)}^{x}}-{{2}^{3x-2}}\times {{\left( 2\times 2 \right)}^{x+1}}}{3\times 5\times {{2}^{x-1}}\times {{\left( 2\times 2\times 2\times 2 \right)}^{x}}}-\dfrac{5\cdot {{5}^{x-1}}}{\sqrt{{{5}^{2x}}}} \\
 & \Rightarrow \dfrac{{{2}^{4}}\cdot {{\left( {{2}^{5}} \right)}^{x}}-{{2}^{3x-2}}\times {{\left( {{2}^{2}} \right)}^{x+1}}}{3\times 5\times {{2}^{x-1}}\times {{\left( {{2}^{4}} \right)}^{x}}}-\dfrac{5\cdot {{5}^{x-1}}}{{{\left( {{5}^{2x}} \right)}^{\dfrac{1}{2}}}} \\
\end{align}\]
Let us use the exponential identity ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ for $a=2,m=4,n=x$ in ${{\left( {{2}^{4}} \right)}^{x}}$, $a=2,m=2,n=x+1$in ${{\left( {{2}^{2}} \right)}^{x+1}}$ in the numerator of first term. We use the same identity in the denominator of first term for $a=2,m=4,n=x$in ${{\left( {{2}^{4}} \right)}^{x}}$ and also in the denominator of second term $a=2,m=2x,n=\dfrac{1}{2}$in ${{\left( {{5}^{2x}} \right)}^{\dfrac{1}{2}}}$. We proceed,
\[\begin{align}
  & \Rightarrow \dfrac{{{2}^{4}}\cdot {{2}^{5\times x}}-{{2}^{3x-2}}\times {{2}^{2\left( x+1 \right)}}}{3\times 5\times {{2}^{x-1}}\times {{2}^{4\left( x \right)}}}-\dfrac{5\cdot {{5}^{x-1}}}{{{5}^{2x\times \dfrac{1}{2}}}} \\
 & \Rightarrow \dfrac{{{2}^{4}}\cdot {{2}^{5x}}-{{2}^{3x-2}}\times {{2}^{2x+2}}}{3\times 5\times {{2}^{x-1}}\times {{2}^{4x}}}-\dfrac{5\cdot {{5}^{x-1}}}{{{5}^{x}}} \\
\end{align}\]
We use the law of exponentiation ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ for the common base $a=2$ in the numerator and denominator of first term and common base $a=5$ in the numerator of the second term. We proceed to have,
\[\begin{align}
  & \Rightarrow \dfrac{{{2}^{4+5x}}-{{2}^{3x-2+2x+2}}}{3\times 5\times {{2}^{x-1+4x}}}-\dfrac{{{5}^{1+x-1}}}{{{5}^{x}}} \\
 & \Rightarrow \dfrac{{{2}^{4+5x}}-{{2}^{5x}}}{3\times 5\times {{2}^{5x-1}}}-\dfrac{{{5}^{x}}}{{{5}^{x}}} \\
 & \Rightarrow \dfrac{{{2}^{4+5x}}-{{2}^{5x}}}{3\times 5\times {{2}^{5x-1}}}-1 \\
\end{align}\]
We take ${{2}^{5x}}$ common from the numerator of the first term using ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ in ${{2}^{4+5x}}$. We use the law of exponentiation of division with same base ${{a}^{m-n}}=\dfrac{{{a}^{m}}}{{{a}^{n}}}$with base $a=2,m=5x,n=1$ in the denominator of first term. We proceed to have,
\[\begin{align}
  & \Rightarrow \dfrac{{{2}^{4}}\cdot {{2}^{5x}}-{{2}^{5x}}}{3\times 5\times {{2}^{5x-1}}}-1 \\
 & \Rightarrow \dfrac{{{2}^{5x}}\left( {{2}^{4}}-1 \right)}{3\times 5\times \dfrac{{{2}^{5x}}}{2}}-1 \\
 & \Rightarrow \dfrac{{{2}^{5x}}\left( 16-1 \right)}{3\times 5\times {{2}^{5x}}}\times 2-1 \\
 & \Rightarrow \dfrac{{{2}^{5x}}\times 15}{3\times 5\times {{2}^{5x}}}\times 2-1 \\
\end{align}\]
We factorize 15 in the numerator and then cancel out the same terms to have,
\[\begin{align}
  & \Rightarrow \dfrac{{{2}^{5x}}\times 3\times 5}{3\times 5\times {{2}^{5x}}}\times 2-1 \\
 & \Rightarrow 2-1=1 \\
 & \therefore \dfrac{16{{\left( 32 \right)}^{x}}-{{2}^{3x-2}}\times {{4}^{x+1}}}{15\times {{2}^{x-1}}\times {{16}^{x}}}-\dfrac{5\cdot {{5}^{x-1}}}{\sqrt{{{5}^{2x}}}}=1 \\
\end{align}\]
So the correct option is A.

Note: We need to be careful when we add or subtract the powers. We also must note that when we say cancelled out it means we are dividing numerator and denominator with the same terms. If ${{p}^{\text{th}}}$ root of a number $a$ can be written as $a$ raised to the power $p$ which means $\sqrt[p]{a}={{a}^{\dfrac{1}{p}}}$ , we have used for $\sqrt{{{5}^{2x}}}$ in this problem. The prime factorization of a composite number $n$ with $n$ prime factors ${{p}_{1}},{{p}_{2}},...,{{p}_{m}}$ can be written as $n={{p}_{1}}^{{{e}_{1}}}{{p}_{2}}^{{{e}_{2}}}...{{p}_{m}}^{{{e}_{m}}}$ where ${{e}_{1}},{{e}_{2}},...{{e}_{m}}$are positive integral exponents.