# What is the value of x in the equation (a/b)^(2x-3) = (b^3/a^3)^(x+4)?

Aug 11, 2018

$x = - \frac{9}{5}$

#### Explanation:

Provided information is

${\left(\frac{a}{b}\right)}^{2 x - 3} = {\left({b}^{3} / {a}^{3}\right)}^{x + 4}$

so we can rearrange it as,

${\left(\frac{a}{b}\right)}^{2 x - 3} = {\left(\frac{b}{a}\right)}^{3 \left(x + 4\right)} = {\left(\frac{b}{a}\right)}^{3 x + 12} = {\left(\frac{a}{b}\right)}^{- 3 x - 12}$

{ as, $\frac{1}{b} ^ \left(- x\right) = {b}^{0} / {b}^{-} x = {b}^{0 - \left(- x\right)} = {b}^{x}$ similarly, $\frac{1}{a} ^ x$ can be written as ${a}^{0} / {a}^{x} = {a}^{0 - x} = {a}^{-} x$}

Now,comparing,both sides,

$2 x - 3 = - 3 x - 12$

or, $5 x = - 9$

or, $x = - \frac{9}{5}$

Aug 11, 2018

$x = - \frac{9}{5}$

#### Explanation:

${\left(\frac{a}{b}\right)}^{2 x - 3} = {\left({b}^{3} / {a}^{3}\right)}^{x + 4}$

Recall;

$\textcolor{\in \mathrm{di} g o}{\left({x}^{a} / {y}^{a}\right) = {\left(\frac{x}{y}\right)}^{a}}$

Hence;

(a/b)^(2x-3) = (b/a)^(3(x+4)

Simplifying;

${\left(\frac{a}{b}\right)}^{2 x - 3} = {\left(\frac{b}{a}\right)}^{3 x + 12}$

Also recall;

$\textcolor{\in \mathrm{di} g o}{\frac{y}{x} = \frac{1}{\frac{x}{y}} \mathmr{and} {\left(\frac{x}{y}\right)}^{-} 1}$

Hence;

${\left(\frac{a}{b}\right)}^{2 x - 3} = \frac{1}{\frac{a}{b}} ^ \left(3 x + 12\right)$

Simplifying;

(a/b)^(2x-3) = (a/b)^(-1(3x+12)

${\left(\frac{a}{b}\right)}^{2 x - 3} = {\left(\frac{a}{b}\right)}^{- 3 x - 12}$

${\cancel{\left(\frac{a}{b}\right)}}^{2 x - 3} = {\cancel{\left(\frac{a}{b}\right)}}^{- 3 x - 12}$

$2 x - 3 = - 3 x - 12$

Collecting like terms;

$2 x + 3 x = - 12 + 3$

$5 x = - 9$

$x = - \frac{9}{5}$

Aug 11, 2018

${\left(\frac{a}{b}\right)}^{2 x - 3} = {\left({b}^{3} / {a}^{3}\right)}^{x + 4}$

$\implies {a}^{2 x - 3} \times {a}^{3 x + 12} = {b}^{3 x + 12} \times {b}^{2 x - 3}$

$\implies {a}^{5 x + 9} = {b}^{5 x + 9}$

$\implies {a}^{5 x + 9} / {b}^{5 x + 9} = 1$

$\implies {\left(\frac{a}{b}\right)}^{5 x + 9} = {\left(\frac{a}{b}\right)}^{0}$

$\implies 5 x + 9 = 0$

$\implies x = - \frac{9}{5}$