Question

Simplify and write in exponential form with a positive exponent
$\dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}}$ .

Answer 1

Hint: To solve this type of problem, we should know about the Laws of Exponents. First, we should convert the negative exponents into positive exponents and then simplify the problem. This can be done by using the Laws of Exponents. The formulae that we will be using in this problem are given in the following formula section. After making all the exponents positive we will simplify it (if we can) then we will express it in the form where all the exponents are positive.

Complete step by step solution:
We aim to simplify the given expression and give the answer in an exponential form where all the exponents are positive.
Take this Expression
$\dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}}$
We know that the cube of two equals eight. This equation can be written as
$\dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}} = \dfrac{{{{({2^3})}^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}} \xrightarrow[{\begin{array}{*{20}{c}}
{}&{}
\end{array}}]{} (1)$
Now using the formula ${({a^m})^n} = {a^{m \times n}}$ we can write the numerator of the above expression as ${({2^3})^{ - 1}} = {2^3}^{ \times - 1}$
$ \Rightarrow {2^{ - 3}}$
Substitute the above value in Equation $(1)$
$\dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}} = \dfrac{{{2^{ - 3}} \times {5^3}}}{{{2^{ - 4}}}}$
This can be written as
\[\dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}} = \dfrac{{{2^{ - 3}}}}{{{2^{ - 4}}}} \times {5^3} \xrightarrow{{\begin{array}{*{20}{c}}
{}&{}
\end{array}}} (2)\]
$\dfrac{{{2^{ - 3}}}}{{{2^{ - 4}}}}$ has the same base $2$ but a different exponent value.
By using the Quotient Rule (If both the Base values are the same, then we can Subtract the Powers)
$
\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}} \\
\\
\dfrac{{{2^{ - 3}}}}{{{2^{ - 4}}}} = {2^{ - 3 - ( - 4)}} \\
$
$
= {2^{ - 3 + 4}} \\
= {2^1} = 2 \\
$
When the above Value is substituted in the equation $(2)$ , we get
$\dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}} = 2 \times {5^3}$.

Note:
We can also solve this problem by using the negative exponent rule, that is if the power is Negative, then we can reciprocate and change the sign of the negative power to a positive power.
i.e., ${a^{ - m}} = \dfrac{1}{{{a^m}}}$
where $a$ is the Base Value and $m$ is the exponent part
Equation $(2)$ is $\dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}} = \dfrac{{{2^{ - 3}}}}{{{2^{ - 4}}}} \times {5^3}$
$\dfrac{1}{{{2^{ - 4}}}}$ can be written as ${2^4}$
$\dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}} = {2^{ - 3}} \times {2^4} \times {5^3}$
Apply Multiplication rule that is if both the Base values are the same, then we can add the Powers
${a^m} \times {a^n} = {a^{m + n}}$ where $a$ is the Base Value and $m\& n$ are the exponent part
$\dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}} = {2^{ - 3 + 4}} \times {5^3}$
$ \Rightarrow = 2 \times {5^3}$