Question

Express ${\left[ {{{\left( {{2^{\dfrac{1}{2}}} \cdot {4^{\dfrac{3}{4}}} \cdot {8^{\dfrac{5}{6}}} \cdot {{16}^{\dfrac{7}{8}}} \cdot {{32}^{\dfrac{9}{{10}}}}} \right)}^4}} \right]^{\dfrac{8}{{25}}}}$ as an integer
A) 8192
B) 32768
C) 65536
D) 131072

Answer 1

Hint:
Firstly, we need to simplify ${\left[ {{{\left( {{2^{\dfrac{1}{2}}} \cdot {4^{\dfrac{3}{4}}} \cdot {8^{\dfrac{5}{6}}} \cdot {{16}^{\dfrac{7}{8}}} \cdot {{32}^{\dfrac{9}{{10}}}}} \right)}^4}} \right]^{\dfrac{8}{{25}}}}$ into simple ${2^x}$ form.
For that, put the values $4 = {2^2},8 = {2^3},16 = {2^4},32 = {2^5}$.
Then, for further simplification, use the properties ${a^x} \cdot {a^y} = {a^{x + y}}$ and ${\left( {{a^x}} \right)^y} = {a^{x \times y}}$ and find the simple form ${2^x}$ .
Thus, multiply 2 x times to get integer form.

Complete step by step solution:
We are asked to find the value of ${\left[ {{{\left( {{2^{\dfrac{1}{2}}} \cdot {4^{\dfrac{3}{4}}} \cdot {8^{\dfrac{5}{6}}} \cdot {{16}^{\dfrac{7}{8}}} \cdot {{32}^{\dfrac{9}{{10}}}}} \right)}^4}} \right]^{\dfrac{8}{{25}}}}$ in integer.
Firstly, we can write the values $4 = {2^2},8 = {2^3},16 = {2^4},32 = {2^5}$ .
So, ${\left[ {{{\left( {{2^{\dfrac{1}{2}}} \cdot {4^{\dfrac{3}{4}}} \cdot {8^{\dfrac{5}{6}}} \cdot {{16}^{\dfrac{7}{8}}} \cdot {{32}^{\dfrac{9}{{10}}}}} \right)}^4}} \right]^{\dfrac{8}{{25}}}} = {\left[ {{{\left( {{2^{\dfrac{1}{2}}} \cdot {2^{2 \times \dfrac{3}{4}}} \cdot {2^{3 \times \dfrac{5}{6}}} \cdot {2^{4 \times \dfrac{7}{8}}} \cdot {2^{5 \times \dfrac{9}{{10}}}}} \right)}^4}} \right]^{\dfrac{8}{{25}}}}$
    \[ = {\left[ {{{\left( {{2^{\dfrac{1}{2}}} \cdot {2^{\dfrac{3}{2}}} \cdot {2^{\dfrac{5}{2}}} \cdot {2^{\dfrac{7}{2}}} \cdot {2^{\dfrac{9}{2}}}} \right)}^4}} \right]^{\dfrac{8}{{25}}}}\]
Now, on applying the property ${a^x} \cdot {a^y} = {a^{x + y}}$ , we get
 \[{\left[ {{{\left( {{2^{\dfrac{1}{2} + \dfrac{3}{2} + \dfrac{5}{2} + \dfrac{7}{2} + \dfrac{9}{2}}}} \right)}^4}} \right]^{\dfrac{8}{{25}}}}\]
 $
   = {\left[ {{{\left( {{2^{\dfrac{{1 + 3 + 5 + 7 + 9}}{2}}}} \right)}^4}} \right]^{\dfrac{8}{{25}}}} \\
   = {\left[ {{{\left( {{2^{\dfrac{{25}}{2}}}} \right)}^4}} \right]^{\dfrac{8}{{25}}}} \\
 $
Now, we will apply the property ${\left( {{a^x}} \right)^y} = {a^{x \times y}}$ on the above value
 $
  {2^{\dfrac{{25}}{2} \times 4 \times \dfrac{8}{{25}}}} \\
   = {2^{16}} \\
 $
Thus, we get ${2^{16}}$ on simplifying ${\left[ {{{\left( {{2^{\dfrac{1}{2}}} \cdot {4^{\dfrac{3}{4}}} \cdot {8^{\dfrac{5}{6}}} \cdot {{16}^{\dfrac{7}{8}}} \cdot {{32}^{\dfrac{9}{{10}}}}} \right)}^4}} \right]^{\dfrac{8}{{25}}}}$ .
Now, we need to convert ${2^{16}}$ in integer form. Thus, ${2^{16}} = 65536$ .
Thus, ${\left[ {{{\left( {{2^{\dfrac{1}{2}}} \cdot {4^{\dfrac{3}{4}}} \cdot {8^{\dfrac{5}{6}}} \cdot {{16}^{\dfrac{7}{8}}} \cdot {{32}^{\dfrac{9}{{10}}}}} \right)}^4}} \right]^{\dfrac{8}{{25}}}} = 65536$.

Note:
Some properties of exponents:
 $
  {a^n} \cdot {a^m} = {a^{n + m}} \\
  {a^n} \cdot {b^n} = {\left( {a \cdot b} \right)^n} \\
  \dfrac{{{a^n}}}{{{a^m}}} = {a^{n - m}} \\
  \dfrac{{{a^n}}}{{{b^n}}} = {\left( {\dfrac{a}{b}} \right)^n} \\
  {\left( {{b^n}} \right)^m} = {b^{nm}} \\
  \sqrt[m]{{{b^n}}} = {b^{\dfrac{n}{m}}} \\
  \sqrt[n]{b} = {b^{\dfrac{1}{n}}} \\
  {b^{ - n}} = \dfrac{1}{{{b^n}}} \\
 $