Question

Find the value of $\dfrac{{{{(0.6)}^0} - {{(0.1)}^{ - 1}}}}{{{{\left( {\dfrac{3}{8}} \right)}^{ - 1}}{{\left( {\dfrac{3}{2}} \right)}^3} + {{\left( {\dfrac{{ - 1}}{3}} \right)}^{ - 1}}}}$
A) $\dfrac{{ - 2}}{3}$
B) $\dfrac{3}{2}$
C) $ - \dfrac{3}{2}$
D) $\dfrac{2}{3}$

Answer 1

Hint:
In the above problem, we will first convert the decimal values to fractions. Then we will apply appropriate laws of exponents to the numerator and denominator. Finally, we will simplify the resulting fraction to get the required value.

Formula used:
We will use laws of exponents:
1) ${(a)^0} = 1$
2) ${(a)^{ - 1}} = \dfrac{1}{a}$
3) $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}},m > n$
4) ${\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}$

Complete step by step solution:
The given expression is $\dfrac{{{{(0.6)}^0} - {{(0.1)}^{ - 1}}}}{{{{\left( {\dfrac{3}{8}} \right)}^{ - 1}}{{\left( {\dfrac{3}{2}} \right)}^3} + {{\left( {\dfrac{{ - 1}}{3}} \right)}^{ - 1}}}}$ …………………………..$\left( 1 \right)$
 We will first write $0.6$ and $0.1$ as fractional values.
We can write $0.6$ as $\dfrac{6}{{10}}$ and $0.1$ as $\dfrac{1}{{10}}$.
The numerator becomes ${\left( {\dfrac{6}{{10}}} \right)^0} - {\left( {\dfrac{1}{{10}}} \right)^{ - 1}}$.
We will now apply the laws of exponents ${(a)^0} = 1$ and ${(a)^{ - 1}} = \dfrac{1}{a}$ to ${\left( {\dfrac{6}{{10}}} \right)^0}{\text{ and }}{\left( {\dfrac{1}{{10}}} \right)^{ - 1}}$ respectively.
Thus, we get
${\left( {\dfrac{6}{{10}}} \right)^0} - {\left( {\dfrac{1}{{10}}} \right)^{ - 1}} = 1 - \left( {\dfrac{1}{{\left( {\dfrac{1}{{10}}} \right)}}} \right)$
Simplifying the expression, we get
$ \Rightarrow {\left( {\dfrac{6}{{10}}} \right)^0} - {\left( {\dfrac{1}{{10}}} \right)^{ - 1}} = 1 - 10 = - 9$
So, the numerator’s value is $ - 9$.
Now, the denominator is ${\left( {\dfrac{3}{8}} \right)^{ - 1}}{\left( {\dfrac{3}{2}} \right)^3} + {\left( {\dfrac{{ - 1}}{3}} \right)^{ - 1}}$.
Applying ${(a)^{ - 1}} = \dfrac{1}{a}$ to ${\left( {\dfrac{3}{8}} \right)^{ - 1}}$ , we get
${\left( {\dfrac{3}{8}} \right)^{ - 1}} = \left( {\dfrac{1}{{\left( {\dfrac{3}{8}} \right)}}} \right) = \dfrac{8}{3}$.
 Again, applying ${(a)^{ - 1}} = \dfrac{1}{a}$ to ${\left( {\dfrac{{ - 1}}{3}} \right)^{ - 1}}$, we have
${\left( {\dfrac{{ - 1}}{3}} \right)^{ - 1}} = \left( {\dfrac{1}{{\left( {\dfrac{{ - 1}}{3}} \right)}}} \right) = \dfrac{3}{{ - 1}} = - 3$.
Thus,
 ${\left( {\dfrac{3}{8}} \right)^{ - 1}}{\left( {\dfrac{3}{2}} \right)^3} + {\left( {\dfrac{{ - 1}}{3}} \right)^{ - 1}} = \left( {\dfrac{8}{3}} \right){\left( {\dfrac{3}{2}} \right)^3} + ( - 3)$ ……………………………………$\left( 2 \right)$
 We can write 8 in terms of powers of 2 as $8 = 2 \times 2 \times 2 = {2^3}$. Substituting this in equation $\left( 2 \right)$, we have
 $ \Rightarrow {\left( {\dfrac{3}{8}} \right)^{ - 1}}{\left( {\dfrac{3}{2}} \right)^3} + {\left( {\dfrac{{ - 1}}{3}} \right)^{ - 1}} = \left( {\dfrac{{{2^3}}}{3}} \right){\left( {\dfrac{3}{2}} \right)^3} + ( - 3)$ ……………………………………$\left( 3 \right)$
We will now apply ${\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}$ to ${\left( {\dfrac{3}{2}} \right)^3}$. This gives us
${\left( {\dfrac{3}{2}} \right)^3} = \dfrac{{{3^3}}}{{{2^3}}}$.
Substituting this in equation $\left( 3 \right)$, we get
$ \Rightarrow {\left( {\dfrac{3}{8}} \right)^{ - 1}}{\left( {\dfrac{3}{2}} \right)^3} + {\left( {\dfrac{{ - 1}}{3}} \right)^{ - 1}} = \left( {\dfrac{{{2^3}}}{3}} \right)\left( {\dfrac{{{3^3}}}{{{2^3}}}} \right) + ( - 3)$ …………………………………….$\left( 4 \right)$
We see that in the first term, ${2^3}$ is both in the numerator and in the denominator. We can cancel out ${2^3}$. So, equation $\left( 4 \right)$ becomes
${\left( {\dfrac{3}{8}} \right)^{ - 1}}{\left( {\dfrac{3}{2}} \right)^3} + {\left( {\dfrac{{ - 1}}{3}} \right)^{ - 1}} = \left( {\dfrac{{{3^3}}}{3}} \right) + ( - 3)$…………………………………….$\left( 5 \right)$
Let us now apply the law $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}},m > n$ to $\dfrac{{{3^3}}}{3}$, where $m = 3$and $n = 1$. Hence, we get
${\left( {\dfrac{3}{8}} \right)^{ - 1}}{\left( {\dfrac{3}{2}} \right)^3} + {\left( {\dfrac{{ - 1}}{3}} \right)^{ - 1}} = 9 + ( - 3) = 6$.
Thus, the value of the denominator is 6.
Using the value of the numerator and denominator in equation $\left( 1 \right)$, we have
$\dfrac{{{{(0.6)}^0} - {{(0.1)}^{ - 1}}}}{{{{\left( {\dfrac{3}{8}} \right)}^{ - 1}}{{\left( {\dfrac{3}{2}} \right)}^3} + {{\left( {\dfrac{{ - 1}}{3}} \right)}^{ - 1}}}} = \dfrac{{ - 9}}{6}$.
We have to reduce $\dfrac{{ - 9}}{6}$ to its lowest form. Both $ - 9{\text{ and }}6$are divisible by 3.
So, $\dfrac{{ - 9}}{6} = \dfrac{{ - 3}}{2}$. Therefore,
$ \Rightarrow \dfrac{{{{(0.6)}^0} - {{(0.1)}^{ - 1}}}}{{{{\left( {\dfrac{3}{8}} \right)}^{ - 1}}{{\left( {\dfrac{3}{2}} \right)}^3} + {{\left( {\dfrac{{ - 1}}{3}} \right)}^{ - 1}}}} = \dfrac{{ - 3}}{2}$

Thus option C the correct option.

Note:
The law of exponent ${(a)^{ - 1}} = \dfrac{1}{a}$ is a particular case of ${(a)^{ - m}} = \dfrac{1}{{{a^m}}}$, where $m > 0$. Here we have used $m = 1$. Here, we have converted the decimal number in numerator into fraction because the denominator has numbers in fraction form. Also, it is easy to apply rules of exponent on fraction then on the decimal number.