# Exponential Equation 2^(x+1) + 2^(x+2) + 3(2^(x-2)) = 8(3^(x-2)) ?

## A step by step solution of the following would be greatly appreciated Thank you :D

Jun 4, 2016

$x = 5$

#### Explanation:

Consider ${2}^{x + 1} \to {2}^{x} \times 2$

Consider ${2}^{x + 2} \to {2}^{x} \times {2}^{2}$

Consider $3 \left({2}^{x - 2}\right) \to {2}^{x} \times \frac{3}{{2}^{2}}$

Consider $8 \left({3}^{x - 2}\right) \to {3}^{x} \times \frac{8}{3} ^ 2$

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$\text{ } {2}^{x} \left(2 + {2}^{2} + \frac{3}{2} ^ 2\right) = \frac{8}{3} ^ 2 \times {3}^{x}$

$\text{ } {2}^{x} \times \frac{27}{4} = \frac{8}{9} \times {3}^{x}$

Divide both sides by ${3}^{x}$

$\text{ } {2}^{x} / \left({3}^{x}\right) \times \frac{27}{4} = \frac{8}{9} \times {3}^{x} / \left({3}^{x}\right)$

But $\frac{{3}^{x}}{{3}^{x}} = 1$

$\text{ } {2}^{x} / \left({3}^{x}\right) \times \frac{27}{4} = \frac{8}{9}$

Multiply both sides by $\frac{4}{27}$

$\text{ } {2}^{x} / \left({3}^{x}\right) \times \frac{27}{4} \times \frac{4}{27} = \frac{8}{9} \times \frac{4}{27}$

But $\frac{27}{4} \times \frac{4}{27} = \frac{27}{27} \times \frac{4}{4} = 1$

$\text{ } {2}^{x} / \left({3}^{x}\right) = \frac{8}{9} \times \frac{4}{27}$

$\text{ } {2}^{x} / {3}^{x} = \frac{32}{243}$
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From this both should be true that

$\text{ "2^x=32" and } {3}^{x} = 243$

Using logs

$\text{ } x \ln \left(2\right) = \ln \left(32\right) \implies x = \ln \frac{32}{\ln} \left(2\right)$

$\text{ } \implies x = 5$
'..........................................

" "xln(3)=ln(243)=> x=ln(243)/(ln(3)

$\text{ } \implies x = 5$

Jun 4, 2016

$x = 5$

#### Explanation:

$2 \times {2}^{x} + 4 \times {2}^{x} + \frac{3}{4} \times {2}^{x} = \left(\frac{8}{9}\right) {3}^{x}$

grouping

$\left(\frac{27}{4}\right) {2}^{x} = \left(\frac{8}{9}\right) {3}^{x}$
${\left(\frac{3}{2}\right)}^{x} = \frac{27 \times 9}{8 \times 4} = {\left(\frac{3}{2}\right)}^{5}$

$x = 5$