Question
$\Big(\frac{7}{3}\Big)^{2}\times\Big(\frac{7}{3}\Big)^{5}=\Big(\frac{7}{3}\Big)^{10}$

Answer

False.
Solution:
Here,$\Big(\frac{7}{3}\Big)^{2}\times\Big(\frac{7}{3}\Big)^{5}=\Big(\frac{7}{3}\Big)^{2+5}$
$=\Big(\frac{7}{3}\Big)^{7}$ $\big[\because\text{a}^{\text{m}}\times\text{a}^{\text{n}}=\text{a}^{\text{m+n}}\big]$
Here, $\Big(\frac{7}{3}\Big)^{2}\times\Big(\frac{7}{3}\Big)^{5}\neq\Big(\frac{7}{3}\Big)^{10}$

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