Question

{NO SPAMMING} only for genius @spammers stay away
answer should be explained and correct
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Answer 1

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

Given:

• $\left(\dfrac{2}{3}\right)^{10} \times \left[\left(\dfrac{3}{2}\right)^2\right]^5 = \left(\dfrac{2}{3}\right)^{2n-2}$

To Find:

• Value of n

Solution:

$\left(\dfrac{2}{3}\right)^{10} \times \left[\left(\dfrac{3}{2}\right)^2\right]^5 = \left(\dfrac{2}{3}\right)^{2n-2} \\\\\implies \left(\dfrac{2}{3}\right)^{10} \times \left(\dfrac{3}{2}\right)^{10} = \left(\dfrac{2}{3}\right)^{2n-2} \\\\\implies \left(\dfrac{2}{3}\right)^{10} \times \left(\dfrac{2}{3}\right)^{-10} = \left(\dfrac{2}{3}\right)^{2n-2} \\\\\implies \left(\dfrac{2}{3}\right)^{10-10} = \left(\dfrac{2}{3}\right)^{2n-2} \\\\\implies \left(\dfrac{2}{3}\right)^{0} = \left(\dfrac{2}{3}\right)^{2n-2}</p><p>\\\\\text{As bases are same, we compare the powers} \\\\\implies 0 = 2n - 2 \\\\\implies 2n = 2 \\\\\bf{\implies n = 1}$

Formula Used:

• $a^m \times a^n= a^{m+n}$
• $\dfrac{1}{a^n} = a^{-n}$
• $(a^m)^n = a^{mn}$

Learn More:

$\begin{gathered}\boxed{\begin{minipage}{5 cm}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{minipage}}\end{gathered}$

Hope it helps!!