How do you solve the exponential equation $8^{x - 1} = 2^{x + 2}$ $8^(x-1)=2^(x+2)$ ?

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How do you solve the exponential equation $8^{x - 1} = 2^{x + 2}$ $8^(x-1)=2^(x+2)$ ?

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Answer 1 · 2017-06-25T12:49:56

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

Explanation:

$express the left side in base 2$ $"express the left side in "color(blue)"base 2"$

$\Rightarrow {(2^{3})}^{x - 1} = 2^{x + 2}$ $rArr(2^3)^(x-1)=2^(x+2)$

$\Rightarrow 2^{(3 x - 3)} = 2^{(x + 2)}$ $rArr2^((3x-3))=2^((x+2))$

$since the bases are equal we can equate the exponents$ $"since the bases are equal we can equate the exponents"$

$\Rightarrow 3 x - 3 = x + 2$ $rArr3x-3=x+2$

$\Rightarrow 2 x = 5$ $rArr2x=5$

$\Rightarrow x = \frac{5}{2}$ $rArrx=5/2$

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How do you solve the exponential equation $8^{x - 1} = 2^{x + 2}$ $8^(x-1)=2^(x+2)$ ?

1 Answer

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How do you solve the exponential equation $8^{x - 1} = 2^{x + 2}$ $8^(x-1)=2^(x+2)$ ?