Exponential Equation $2^{x + 1} + 2^{x + 2} + 3 (2^{x - 2}) = 8 (3^{x - 2})$ $2^(x+1) + 2^(x+2) + 3(2^(x-2)) = 8(3^(x-2))$ ?

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Exponential Equation $2^{x + 1} + 2^{x + 2} + 3 (2^{x - 2}) = 8 (3^{x - 2})$ $2^(x+1) + 2^(x+2) + 3(2^(x-2)) = 8(3^(x-2))$ ?

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Answer 1 · 2016-06-04T16:15:47

$x = 5$ $x=5$

Explanation:

Consider $2^{x + 1} \to 2^{x} \times 2$ $2^(x+1) -> 2^(x)xx2$

Consider $2^{x + 2} \to 2^{x} \times 2^{2}$ $2^(x+2)->2^(x)xx2^2$

Consider $3 (2^{x - 2}) \to 2^{x} \times \frac{3}{2^{2}}$ $3(2^(x-2)) -> 2^(x)xx3/(2^2)$

Consider $8 (3^{x - 2}) \to 3^{x} \times \frac{8}{3^{2}}$ $8(3^(x-2)) ->3^(x)xx8/3^2$

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$2^{x} (2 + 2^{2} + \frac{3}{2^{2}}) = \frac{8}{3^{2}} \times 3^{x}$ $" "2^x(2+2^2+3/2^2) = 8/3^2xx3^x$

$2^{x} \times \frac{27}{4} = \frac{8}{9} \times 3^{x}$ $" "2^(x)xx27/4=8/9xx3^x$

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

$\frac{2^{x}}{3^{x}} \times \frac{27}{4} = \frac{8}{9} \times \frac{3^{x}}{3^{x}}$ $" "2^(x)/(3^x)xx27/4=8/9xx3^x/(3^x)$

But $\frac{3^{x}}{3^{x}} = 1$ $(3^x)/(3^x)=1$

$\frac{2^{x}}{3^{x}} \times \frac{27}{4} = \frac{8}{9}$ $" "2^(x)/(3^x)xx27/4=8/9$

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

$\frac{2^{x}}{3^{x}} \times \frac{27}{4} \times \frac{4}{27} = \frac{8}{9} \times \frac{4}{27}$ $" "2^(x)/(3^x)xx27/4xx4/27=8/9xx4/27$

But $\frac{27}{4} \times \frac{4}{27} = \frac{27}{27} \times \frac{4}{4} = 1$ $27/4xx4/27 =27/27xx4/4 = 1$

$\frac{2^{x}}{3^{x}} = \frac{8}{9} \times \frac{4}{27}$ $" "2^(x)/(3^x)=8/9xx4/27$

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

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

Using logs

$x ln (2) = ln (32) \Rightarrow x = \frac{ln (32)}{ln (2)}$ $" "xln(2)=ln(32) => x= ln(32)/ln(2)$

$\Rightarrow x = 5$ $" "=> x=5$
'..........................................

$x ln (3) = ln (243) \Rightarrow x = \frac{ln (243)}{ln (3)}$ $" "xln(3)=ln(243)=> x=ln(243)/(ln(3)$

$\Rightarrow x = 5$ $" "=>x= 5$

Answer 2 · 2016-06-04T16:34:58

$x = 5$ $x=5$

Explanation:

$2 \times 2^{x} + 4 \times 2^{x} + \frac{3}{4} \times 2^{x} = (\frac{8}{9}) 3^{x}$ $2times 2^x+4 times2^x+3/4 times 2^x=(8/9)3^x$

grouping

$(\frac{27}{4}) 2^{x} = (\frac{8}{9}) 3^{x}$ $(27/4)2^x=(8/9) 3^x$
${(\frac{3}{2})}^{x} = \frac{27 \times 9}{8 \times 4} = {(\frac{3}{2})}^{5}$ $(3/2)^x=(27times9)/(8times 4)=(3/2)^5$

$x = 5$ $x = 5$

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Exponential Equation $2^{x + 1} + 2^{x + 2} + 3 (2^{x - 2}) = 8 (3^{x - 2})$ $2^(x+1) + 2^(x+2) + 3(2^(x-2)) = 8(3^(x-2))$ ?

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Explanation:

Explanation:

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

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2 Answers

Explanation:

Explanation:

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Exponential Equation $2^{x + 1} + 2^{x + 2} + 3 (2^{x - 2}) = 8 (3^{x - 2})$ $2^(x+1) + 2^(x+2) + 3(2^(x-2)) = 8(3^(x-2))$ ?

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