How do you solve $3^{2 x - 1} = 5^{2 - 3 x}$ $3^(2x−1) = 5^(2−3x)$ ?

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How do you solve $3^{2 x - 1} = 5^{2 - 3 x}$ $3^(2x−1) = 5^(2−3x)$ ?

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Answer 1 · 2015-11-28T00:23:39

I found: $x = 0.61454$ $x=0.61454$

Explanation:

Take the natural log of both sides:
${ln 3}^{2 x - 1} = {ln 5}^{2 - 3 x}$ $ln3^(2x-1)=ln5^(2-3x)$
use the property of logs:
${log x}^{a} = a log x$ $logx^a=alogx$
$(2 x - 1) ln 3 = (2 - 3 x) ln 5$ $(2x-1)ln3=(2-3x)ln5$
$2 (ln 3) x - ln 3 = 2 ln 5 - 3 (ln 5) x$ $2(ln3)x-ln3=2ln5-3(ln5)x$
collect $x$ $x$ :
$x [2 (ln 3) + 3 (ln 5)] = 2 ln 5 + ln 3$ $x[2(ln3)+3(ln5)]=2ln5+ln3$
$x [7.02554] = 4.31748$ $x[7.02554]=4.31748$
$x = \frac{4.31748}{7.02554} = 0.61454$ $x=4.31748/7.02554=0.61454$

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How do you solve $3^{2 x - 1} = 5^{2 - 3 x}$ $3^(2x−1) = 5^(2−3x)$ ?

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