Q64. - Algebra 1

Question

For any nonzero real numbers a and b and any integers m and t, simplify the expression ${\left(-\frac{a^m}{b^t}\right)}^{2t}$ and describe each step.

Step-by-Step Solution

Answer

${\left(-\frac{a^m}{b^t}\right)}^{2t}=-\frac{a^{2mt}}{b^{2t^2}}$

Step 1. State the ‘Law of indices’ for dividing powers with same exponent.

When two variables with different bases, but same indices are divided, we are required to divide the bases and raise the same index to it.

${\left(\frac{a}{b}\right)}^p=\frac{a^p}{b^p}$

Step 2. State the ‘Law of indices’ for power of index numbers.

When a variable with some index is again raised with different index, then both the indices are multiplied together raised to the power of the same base.

${\left(a^p\right)}^q=a^{pq}$

Step 3. Simplify the given expression.

Rewrite the given expression.

${\left(-\frac{a^m}{b^t}\right)}^{2t}={\left(\frac{-a^m}{b^t}\right)}^{2t} \left[\text{Using\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\frac{a}{b}=\frac{-a}{b}\right]$

Use the above two rules to simplify the given expression.

Apply rule 7 to the expression ${\left(\frac{-a^m}{b^t}\right)}^{2t}$.

${\left(\frac{-a^m}{b^t}\right)}^{2t}=\frac{{\left(-a^m\right)}^{2t}}{{\left(b^t\right)}^{2t}}$

Now apply rule 5 to the expression width="78" height="52" style="max-width: none;" $\left(\frac{{\left(-a^m\right)}^{2t}}{{\left(b^t\right)}^{2t}}\right)$.

$\begin{array}{c}\frac{{\left(-a^m\right)}^{2t}}{{\left(b^t\right)}^{2t}}=\frac{{\left(-a\right)}^{m\times 2t}}{{\left(b\right)}^{t\times 2t}}\\ \frac{{\left(-a^m\right)}^{2t}}{{\left(b^t\right)}^{2t}}=\frac{{\left(-a\right)}^{2mt}}{{\left(b\right)}^{2t^2}}\end{array}$

Since the power of ‘a’ is 2mt, and 2mt will always be an even number as it’s a multiple of 2. 

Apply the exponent rule ${\left(-a\right)}^n=a^n$, if n is even.

${\left(-a\right)}^{m\times 2t}=a^{2mt}$

Therefore, the simplified expression is as follows:

$\begin{array}{l}\frac{{\left(-a^m\right)}^{2t}}{{\left(b^t\right)}^{2t}}=\frac{{\left(a\right)}^{2mt}}{{\left(b\right)}^{2t^2}}\\ \frac{{\left(-a^m\right)}^{2t}}{{\left(b^t\right)}^{2t}}=\frac{a^{2mt}}{b^{2t^2}}\end{array}$

Therefore ${\left(-\frac{a^m}{b^t}\right)}^{2t}=\frac{a^{2mt}}{b^{2t^2}}$.