Question

Find the value of $(\frac{4}{49}{)}^{\frac{3}{4}}\times (\frac{49}{4}{)}^{-\frac{1}{2}}.$

• A. $\frac{4}{49}$
  
• B. $\frac{49}{4}$
  
• C. $(\frac{4}{49}{)}^{\frac{5}{4}}$
  
• D. $(\frac{49}{4}{)}^{\frac{5}{4}}$
  

Answer 1

The correct option is C $(\frac{4}{49}{)}^{\frac{5}{4}}$


$\underset{–}{\text{Need to simplify:}}$ $(\frac{4}{49}{)}^{\frac{3}{4}}\times (\frac{49}{4}{)}^{-\frac{1}{2}}$

In this case, no exponent rule is directly applicable. So, let's try to make bases equal.

$(\frac{4}{49}{)}^{\frac{3}{4}}\times (\frac{49}{4}{)}^{-\frac{1}{2}}$

$=(\frac{4}{49}{)}^{\frac{3}{4}}\times (\frac{4}{49}{)}^{\frac{1}{2}}$ $[\text{Applying}(\frac{a}{b}{)}^{-n}=(\frac{b}{a}{)}^n\text{on}(\frac{49}{4}{)}^{-\frac{1}{2}}]$

Now, both the exponents have the same base; thus applying exponent rule: $[a^m\times a^n=a^{m+n}]$

$=(\frac{4}{49}{)}^{\frac{3}{4}+\frac{1}{2}}$

$=(\frac{4}{49}{)}^{\frac{3}{4}+\frac{2}{4}}$

$=(\frac{4}{49}{)}^{\frac{3+2}{4}}$

$\Rightarrow \underset{–}{(\frac{4}{49}{)}^{\frac{3}{4}}\times (\frac{49}{4}{)}^{-\frac{1}{2}}=(\frac{4}{49}{)}^{\frac{5}{4}}}$

Hence, the value of $(\frac{4}{49}{)}^{\frac{3}{4}}\times (\frac{49}{4}{)}^{-\frac{1}{2}}$ is equal to$(\frac{4}{49}{)}^{\frac{5}{4}}.$

Therefore, option (c.) is the correct answer.