# How do you solve 4^(3x+2)times32^(x-2)=8^(3x-4)?

Sep 15, 2016

The soln. is $x = - 3.$

#### Explanation:

${4}^{3 x + 2} \times {32}^{x - 2} = {8}^{3 x - 4}$

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

$\Rightarrow {2}^{6 x + 4} \cdot {2}^{5 x - 10} = {2}^{9 x - 12} \ldots \ldots \ldots \left[{\left({a}^{m}\right)}^{n} = {a}^{m n}\right]$

$\Rightarrow {2}^{6 x + 4 + 5 x - 10} = {2}^{9 x - 12} \ldots \ldots \ldots \ldots \ldots \left[{a}^{m} \cdot {a}^{n} = {a}^{m + n}\right]$

$\Rightarrow {2}^{11 x - 6} = {2}^{9 x - 12} \ldots \ldots \ldots \ldots . . \left[{a}^{m} = {a}^{n} \Rightarrow m = n\right]$

$\Rightarrow 11 x - 6 = 9 x - 12$

$\Rightarrow 11 x - 9 x = 6 - 12$

$\Rightarrow 2 x = - 6$

$\Rightarrow x = - 3$

This root satisfy the eqn.

Hence, the soln. is $x = - 3.$

Sep 15, 2016

$x = - 3$

#### Explanation:

We have: ${4}^{3 x + 2} \times {32}^{x - 2} = {8}^{3 x - 4}$

Let's express the numbers in terms of $2$:

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

Using the laws of exponents:

$\implies {2}^{6 x + 4} \times {2}^{5 x - 10} = {2}^{9 x - 12}$

$\implies {2}^{6 x + 4 + 5 x - 10} = {2}^{9 x - 12}$

$\implies {2}^{11 x - 6} = {2}^{9 x - 12}$

$\implies 11 x - 6 = 9 x - 12$

$\implies 2 x = - 6$

$\implies x = - 3$

Therefore, the solution to the equation is $x = - 3$.