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

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

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Answer 1 · 2016-01-23T01:24:19

$x \approx 1.7093$ $xapprox1.7093$

Explanation:

$2^{x + 1} = 3^{x}$ $2^(x+1)=3^x$
Taking log of both sides
${log 2}^{x + 1} = {log 3}^{x}$ $log2^(x+1)=log3^x$
$\Rightarrow (x + 1) log 2 = x log 3$ $implies (x+1)log2=xlog3$
$\Rightarrow (x + 1) \times 0.3010 = x \times 0.4771$ $implies (x+1)xx0.3010=x xx0.4771$
Rearranging we get
$(- 0.3010 + 0.4771) x = 0.3010$ $(-0.3010+0.4771)x=0.3010$
$\Rightarrow x = \frac{0.3010}{0.4771 - 0.3010}$ $implies x=0.3010/(0.4771-0.3010)$

$x \approx 1.7093$ $xapprox1.7093$

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

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