Question

Find y in the picture

Answer 1

Laws of Exponents:-

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

$\LARGE\mathfrak{\pink{Solution:-}}$

$\large\mathbb{\orange{GIVEN:-}}$

$\frac{{2}^{(1 - y)} \times {2}^{(y - 1)}}{ {2}^{(y + 2)} } = {8}^{(2 - 3y)}$

$\large\mathbb{\purple{TO FIND:-}}$

Value of y .

$\large\mathbb{\green{FORMULA:-}}$

• $\frac{ {x}^{9} }{ {x}^{5} } = {x}^{(9 - 5)} = {x}^{4}$
• ${ {x}^{9} } \times { {x}^{5} } = {x}^{(9 + 5)} = {x}^{14}$
• ${( {x}^{2} )}^{3} = {x}^{3 \times 2} = {x}^{6}$
• ${a}^{ - 1} = \frac{1}{a}$

$\large\mathbb{\red{ACCORDING \:TO\: THE\: QUESTION:-}}$

$\frac{{2}^{(1 - y)} \times {2}^{(y - 1)}}{ {2}^{(y + 2)} } = {8}^{(2 - 3y)} \\ \\ \implies \: \frac{ {2}^{(1 - y) + (y - 1)} }{ {2}^{(y + 2)} } = {{2}^{3} }^{(2 - 3y)} \\ \\ \implies \: {2}^{(1 - y) + (y - 1) - (y + 2)} = {2}^{6 - 9y} \\ \\ \implies \: {2}^{( - y - 2)} = {2}^{6 - 9y} \\ \\ \implies \: - y - 2 = 6 - 9y\\ \\ \implies \: 9y - y = 6 + 2\\ \\ \implies \: 8y = 8\\ \\ \implies \: y = \cancel\frac{8}{8} \\ \\ \implies \: y \: = 1$

$\large\mathbb{\purple{ANSWER:-}}$

$\\ \implies \boxed{y \: = 1}$

$\large\mathbb{\blue{VERIFICATION:-}}$

$\frac{{2}^{(1 - y)} \times {2}^{(y - 1)}}{ {2}^{(y + 2)} } = {8}^{(2 - 3y)} \\ \\ \implies \: \frac{{2}^{(1 - 1)} \times {2}^{(1 - 1)}}{ {2}^{(1 + 2)} } = {8}^{(2 - (3 \times 1))} \\ \\ \implies \: \frac{ {2}^{0} \times {2}^{0} }{ {2}^{3} } = {8}^{2 - 3} \\ \\ \implies \: \frac{1 \times 1}{8} = {8}^{ - 1} \\ \\ \implies \: \frac{1}{8} = \frac{1}{8} (proved)$