Question

Please solve me this question?​

Answer 1

Answer:

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Answer 2

EXPLANATION:-

$\sf\;(\frac{3}{4})^2\times(\frac{3}{4})^{-4}=(\frac{3}{4})^{3x-2}\\\\Using\;the\;law\;of\;exponen\;a^m\times a^n=a^{m+n}\\\\(\frac{3}{4})^{2+(-4)}=(\frac{3}{4})^{3x-2}\\\\(\frac{3}{4})^{-2}=(\frac{3}{4})^{3x-2}\\\\Cancelling\;\frac{3}{4}\;in\;both\;lhs\;and\;rhs\\\\-2=3x-2\\\\-2+2=3x\\\\0=3x\\\\x=0$

Hence the value of x is 0

EXTRA INFORMATION:-

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