Question

The value of $\dfrac{{{8^{3a}} \times {2^5} \times {2^{2a}}}}{{4 \times {2^{11a}} \times {2^{ - 2a}}}}$ is ${2^{2a + 3}}$. State true or false
A.True
B.False

Answer 1

Hint: In this question, we need to check whether the given exponential function $\dfrac{{{8^{3a}} \times {2^5} \times {2^{2a}}}}{{4 \times {2^{11a}} \times {2^{ - 2a}}}}$ equals to ${2^{2a + 3}}$ or not. For this, we will use different properties of the exponential function and simplify the given function in the simplest form.

Complete step-by-step answer:
The given expression of the function can also be written with the same base as:
$\Rightarrow \dfrac{{{8^{3a}} \times {2^5} \times {2^{2a}}}}{{4 \times {2^{11a}} \times {2^{ - 2a}}}} = \dfrac{{{{\left( {{{(2)}^3}} \right)}^{3a}} \times {2^5} \times {2^{2a}}}}{{{{(2)}^2} \times {2^{11a}} \times {2^{ - 2a}}}} - - - - (i)$
One of the properties of the exponential function states that the raised power of the raised power term of a number can be multiplied together. Mathematically, ${\left( {{a^b}} \right)^c} = {a^{bc}}$.
Now, following the above exponential property in the equation (i), we get
$
\Rightarrow \dfrac{{{8^{3a}} \times {2^5} \times {2^{2a}}}}{{4 \times {2^{11a}} \times {2^{ - 2a}}}} = \dfrac{{{{\left( {{{(2)}^3}} \right)}^{3a}} \times {2^5} \times {2^{2a}}}}{{{{(2)}^2} \times {2^{11a}} \times {2^{ - 2a}}}} \\
   = \dfrac{{{{(2)}^{3 \times 3a}} \times {2^5} \times {2^{2a}}}}{{{{(2)}^2} \times {2^{11a}} \times {2^{ - 2a}}}} - - - - (ii) \\
 $
In the product of the two numbers such that the base of the raised powers of the numbers is the same; then, the raised powers can be added together. Mathematically, ${a^b} \times {a^c} = {a^{\left( {b + c} \right)}}$.
So, applying the above-discussed property of the exponential function in the equation (ii), we get
$
\Rightarrow \dfrac{{{8^{3a}} \times {2^5} \times {2^{2a}}}}{{4 \times {2^{11a}} \times {2^{ - 2a}}}} = \dfrac{{{{(2)}^{3 \times 3a}} \times {2^5} \times {2^{2a}}}}{{{{(2)}^2} \times {2^{11a}} \times {2^{ - 2a}}}} \\
   = \dfrac{{{2^{\left( {9a + 5 + 2a} \right)}}}}{{{2^{\left( {2 + 11a - 2a} \right)}}}} - - - - (iii) \\
 $
Again, using the property of the exponential function, which states that in the division of the two numbers such that the base of the raised powers of the numbers is the same then, the raised powers can be subtracted from each other. Mathematically, \[\dfrac{{{a^b}}}{{{a^c}}} = {a^{\left( {b - c} \right)}}\]
So, applying the above-discussed property of the exponential function in the equation (iii), we get
$
\Rightarrow \dfrac{{{8^{3a}} \times {2^5} \times {2^{2a}}}}{{4 \times {2^{11a}} \times {2^{ - 2a}}}} = \dfrac{{{2^{\left( {9a + 5 + 2a} \right)}}}}{{{2^{\left( {2 + 11a - 2a} \right)}}}} \\
   = {2^{\left[ {\left( {9a + 5 + 2a} \right) - \left( {2 + 11a - 2a} \right)} \right]}} \\
   = {2^{\left[ {\left( {11a + 5} \right) - \left( {2 + 9a} \right)} \right]}} \\
   = {2^{\left( {11a + 5 - 2 - 9a} \right)}} \\
   = {2^{\left( {2a + 3} \right)}} \\
 $
Hence, we can say that $\dfrac{{{8^{3a}} \times {2^5} \times {2^{2a}}}}{{4 \times {2^{11a}} \times {2^{ - 2a}}}} = {2^{2a + 3}}$.
Therefore, the given statement is true.

So, the correct answer is “Option A”.

Note: Students should be careful while using the sign convention in the property of the exponential function. Moreover, we have transformed the base of the numbers to 2 everywhere, as it is the smallest base that we can see in the given expression.