How do you solve ${(\frac{1}{4})}^{2 x} = {(\frac{1}{32})}^{3 x + 1}$ $(1/4)^(2x)=(1/32)^(3x+1)$ ?

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How do you solve ${(\frac{1}{4})}^{2 x} = {(\frac{1}{32})}^{3 x + 1}$ $(1/4)^(2x)=(1/32)^(3x+1)$ ?

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Answer 1 · 2016-05-26T11:21:09

$x = - \frac{5}{11}$ $x = -5/11$

Explanation:

${(\frac{1}{4})}^{2 x} = {(\frac{1}{32})}^{3 x + 1} \equiv \frac{1}{{(4)}^{2 x}} = \frac{1}{{(32)}^{3 x + 1}}$ $(1/4)^{2x}=(1/32)^{3x+1} equiv 1/(4)^{2x}=1/(32)^{3x+1}$
$\frac{1}{{(4)}^{2 x}} = \frac{1}{{(32)}^{3 x + 1}} \equiv {(4)}^{2 x} = {(32)}^{3 x + 1}$ $1/(4)^{2x}=1/(32)^{3x+1} equiv (4)^{2x}=(32)^{3x+1}$
${(4)}^{2 x} = {(32)}^{3 x + 1} \equiv {(2^{2})}^{2 x} = {(2^{5})}^{3 x + 1}$ $(4)^{2x}=(32)^{3x+1} equiv (2^2)^{2x}=(2^5)^{3x+1}$
${(2^{2})}^{2 x} = {(2^{5})}^{3 x + 1} \equiv 2^{2 \times 2 x} = 2^{5 \times (3 x + 1)}$ $(2^2)^{2x}=(2^5)^{3x+1}equiv 2^{2 times 2 x}=2^{5 times(3x+1)}$
$2^{2 \times 2 x} = 2^{5 \times (3 x + 1)} \equiv 4 x = 15 x + 5$ $2^{2 times 2 x}=2^{5 times(3x+1)}equiv 4x=15x+5$
solving for $x$ $x$ we get $x = - \frac{5}{11}$ $x = -5/11$

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How do you solve ${(\frac{1}{4})}^{2 x} = {(\frac{1}{32})}^{3 x + 1}$ $(1/4)^(2x)=(1/32)^(3x+1)$ ?

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