Question

solve the question with whole process in this app

Answer 1

${3}^{x - 1} = {3}^{2} \\ same \: base \: equating \: powers. \\ x - 1 = 2 \\ x = 2 + 1 = 3$
${4}^{y + 2} = {4}^{3} \\ same \: base \: equating \: powers \\ y + 2 = 3 \\ y = 3 - 2 = 1$
$\frac{y}{x} - \frac{x}{y}$
= 1/3 - 3/1
Or, 1/3 - 3
= 1-9/3= 8/3 (taking lcm and solving).

Answer 2

Hello............. ^_^

Here is your answer...

$if \: \: {3}^{x - 1} = 9 \: \: and \: \: {4}^{y + 2} = 64,find \: \: the \: \: value \: \: of \: \: \frac{y}{x} - \frac{x}{y} \\ \\ first \: \: find \: \: value \: \: of \: \: x \: \: in \: \: {3}^{x - 1} = 9 \\ \\ = > {3}^{x - 1} = 9 \\ \\ 9 = {3}^{2} \\ \\ = > {3}^{x - 1} = {3}^{2} \\ now \: \: bases \: \:are \: \: same \: \: \\ now \: \: equating \: \: their \: \: powers \\ = > x - 1 = 2 \\ = > x = 2 + 1 \\ = > x = 3 \\ \\ value \: \: of \: \: x \: = 3 \\ \\ \\ find \: \: the \: \: value \: \: of \: \: y \: \: in \: {4}^{y + 2} = 64 \\ \\ = > {4}^{y + 2} = 64 \\ \\ 64 = {4}^{3} \\ \\ = > {4}^{y + 2} = {4}^{3} \\ \\now \: \: bases \: \: are \: \: same \: \\ now \: equating \: \: their \: \: powers \\ = > y + 2 = 3 \\ = > y = 3 - 2 \\ = > y = 1 \\ \\ value \: \: of \: \: y \: = 1 \\ \\ \\ \\ now \: finding \: \: the \: \: value \: \: of \: \: \frac{y}{x} - \frac{x}{y} \\ \\ = \frac{1}{3} - \frac{3}{1} \\ \\ (take \: \: lcm \: \: 3) \\ \\ = \frac{1 - 9}{1 } \\ \\ = \frac{ - 8}{3}$



Hope it helps........... ^_^