# What is the value of x in 81^(x^3 + 2x^2) = 27^((5*x)/3)?

Jun 6, 2018

$x = \frac{1}{2} \mathmr{and} x = - \frac{5}{2}$

#### Explanation:

${81}^{{x}^{3} + 2 {x}^{2}} = {27}^{\frac{5 x}{3}}$

Note: ${3}^{4} = 81 \mathmr{and} {3}^{3} = 27$

${3}^{4 \left({x}^{3} + 2 {x}^{2}\right)} = {3}^{3 \left(\frac{5 x}{3}\right)}$

${\cancel{3}}^{4 \left({x}^{3} + 2 {x}^{2}\right)} = {\cancel{3}}^{3 \left(\frac{5 x}{3}\right)}$

$4 \left({x}^{3} + 2 {x}^{2}\right) = 3 \left(\frac{5 x}{3}\right)$

$4 \left({x}^{3} + 2 {x}^{2}\right) = \cancel{3} \left(\frac{5 x}{\cancel{3}}\right)$

$4 {x}^{3} + 8 {x}^{2} = 5 x$

Dividing through by $x$

$\frac{4 {x}^{3}}{x} + \frac{8 {x}^{2}}{x} = \frac{5 x}{x}$

$\frac{4 {x}^{{\cancel{3}}^{2}}}{\cancel{x}} + \frac{8 {x}^{{\cancel{2}}^{1}}}{\cancel{x}} = \frac{5 \cancel{x}}{\cancel{x}}$

$4 {x}^{2} + 8 x = 5$

$4 {x}^{2} + 8 x - 5 = 0$

Using Factorisation Method..

$2 \mathmr{and} 10 \to \text{factors}$

Proof: $10 x - 2 x = 8 x \mathmr{and} 10 \times - 2 = - 20$

Therefore;

$4 {x}^{2} - 2 x + 10 x - 5 = 0$

Grouping the factors;

$\left(4 {x}^{2} - 2 x\right) + \left(10 x - 5\right) = 0$

Factorising;

$2 x \left(2 x - 1\right) + 5 \left(2 x - 1\right) = 0$

Seperating the factors;

$\left(2 x - 1\right) \left(2 x + 5\right) = 0$

$2 x - 1 = 0 \mathmr{and} 2 x + 5 = 0$

$2 x = 1 \mathmr{and} 2 x = - 5$

$x = \frac{1}{2} \mathmr{and} x = - \frac{5}{2}$