How do you solve $5^{x - 2} = 3^{3 x + 2}$ $5^(x-2) = 3^(3x+2)$ ?

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How do you solve $5^{x - 2} = 3^{3 x + 2}$ $5^(x-2) = 3^(3x+2)$ ?

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Answer 1 · 2015-12-31T15:14:47

I found $- 3.2116$ $-3.2116$

Explanation:

We can take the natural log of both sides:
${ln (5)}^{x - 2} = {ln (3)}^{3 x + 2}$ $ln(5)^(x-2)=ln(3)^(3x+2)$
We use the property of the logs:
${log x}^{y} = y log x$ $logx^y=ylogx$
$(x - 2) ln (5) = (3 x + 2) ln (3)$ $(x-2)ln(5)=(3x+2)ln(3)$
Rearrange:
$x ln 5 - 2 ln 5 = 3 x ln 3 + 2 ln 3$ $xln5-2ln5=3xln3+2ln3$
$x (ln 5 - 3 ln 3) = 2 ln 5 + 2 ln 3$ $x(ln5-3ln3)=2ln5+2ln3$
$x = \frac{2 ln 5 + 2 ln 3}{ln 5 - 3 ln 3} = - 3.2116$ $x=(2ln5+2ln3)/(ln5-3ln3)=-3.2116$

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How do you solve $5^{x - 2} = 3^{3 x + 2}$ $5^(x-2) = 3^(3x+2)$ ?

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Explanation:

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