Question

Which of the following is not equal to $1$?

• A. $\frac{\left(2^3\times 3^3\right)}{\left(4\times 18\right)}$
• B. $\frac{\left[{\left(-2\right)}^3\times {\left(-2\right)}^4\right]}{{\left(-2\right)}^7}$
• C. $\frac{3^0\times 5^3}{5\times 25}$
• D. $\frac{2^4}{{\left(7^0+3^0\right)}^3}$

Answer 1

Answer: (A), (D)

$\frac{2^4}{{\left(7^0+3^0\right)}^3}$

Step 1. Explanation for the correct option.

Simplify:

$$\begin{gathered} \frac{\left(2^3\times 3^3\right)}{\left(4\times 18\right)}=\frac{\left(8\times 27\right)}{\left(4\times 18\right)} \\ =\frac{216}{72} \\ =3\ne 1 \end{gathered}$$

And,

$$\begin{gathered} \frac{2^4}{{\left(7^0+3^0\right)}^3}=\frac{2^4}{{\left(1+1\right)}^3}...\left[\because a^0=1\right] \\ =\frac{2^4}{{\left(2\right)}^3} \\ =2^{4-3}...\mathbf{[}\mathbf{\because }\frac{{\mathbf{a}}^{\mathbf{m}}}{{\mathbf{a}}^{\mathbf{n}}}\mathbf{=}{\mathbf{a}}^{\mathbf{m}\mathbf{-}\mathbf{n}}\mathbf{]} \\ =2^1 \\ =2 \end{gathered}$$

Therefore, $\frac{\left(2^3\times 3^3\right)}{\left(4\times 18\right)}=3\ne 1$ and $\frac{2^4}{{\left(7^0+3^0\right)}^3}=2\ne 1$

Step 2. Explanation for the incorrect options.

$$\begin{gathered} \frac{\left[{\left(-2\right)}^3\times {\left(-2\right)}^4\right]}{{\left(-2\right)}^7}=\frac{\left[{\left(-2\right)}^{3+4}\right]}{{\left(-2\right)}^7}...\left[\because a^m\times a^n=a^{m+n}\right] \\ =\frac{{\left(-2\right)}^7}{{\left(-2\right)}^7} \\ =1 \end{gathered}$$

And,

$$\begin{gathered} \frac{3^0\times 5^3}{5\times 25}=\frac{1\times 5^3}{5\times 25}...\left[\because a^0=1\right] \\ =\frac{5^3}{5^3} \\ =1 \end{gathered}$$

As, $\frac{\left[{\left(-2\right)}^3\times {\left(-2\right)}^4\right]}{{\left(-2\right)}^7}$ and $\frac{3^0\times 5^3}{5\times 25}$ are equal to $1$.

Hence, the correct options are $\mathrm{A}$ and $\mathrm{D}$.