# How do you solve 2\cdot 3^ { 10x + 6} = 18?

Jul 20, 2018

The solution is $= - \frac{2}{5}$

#### Explanation:

The equation is

$2 \cdot {3}^{10 x + 6} = 18$

Dividing by $2$

${3}^{10 x + 6} = \frac{18}{2} = 9 = {3}^{2}$

Therefore,

The exponents are equal

$10 x + 6 = 2$

$10 x = 2 - 6 = - 4$

Dividing by $10$

$x = - \frac{4}{10} = - \frac{2}{5}$

Jul 20, 2018

$x = - \frac{2}{5}$

#### Explanation:

${3}^{10 x + 6} = 9$

${\log}_{3} {3}^{10 x + 6} = {\log}_{3} 9$

$10 x + 6 = 2$

$10 x = - 4$

$x = - \frac{4}{10}$

$x = - \frac{2}{5}$

Jul 20, 2018

$x = - \frac{2}{5}$

#### Explanation:

Here ,

$2 \cdot {3}^{10 x + 6} = 2 \cdot 9$

$\implies 2 \cdot {3}^{10 x + 6} = 2 \cdot {3}^{2}$

Dividing both sides by $2$

$\frac{\cancel{2} \cdot {3}^{10 x + 6}}{\cancel{2}} = \frac{\cancel{2} \cdot {3}^{2}}{\cancel{2}}$

$\implies {3}^{10 x + 6} = {3}^{2}$

$\implies 10 x + 6 = 2$

Adding both sides $\left(- 6\right)$

$\implies 10 x + 6 + \left(- 6\right) = 2 + \left(- 6\right)$

$\implies 10 x = - 4$

Dividing both sides by $10$

$\implies x = - \frac{4}{10} = - \frac{2 \times 2}{2 \times 5}$

$\implies x = - \frac{2}{5}$

$x = - 0.4$

#### Explanation:

$2 \setminus \cdot {3}^{10 x + 6} = 18$

${3}^{10 x + 6} = \frac{18}{2}$

${3}^{10 x + 6} = 9$

${3}^{10 x + 6} = {3}^{2}$

Comparing the powers on base $3$ on both the sides we get

$10 x + 6 = 2$

$10 x = 2 - 6$

$10 = - 4$

$x = - \frac{4}{10}$

$x = - 0.4$