How do you solve the equation ${0.16}^{4 + 3 x} = {0.3}^{8 - x}$ $0.16^(4+3x)=0.3^(8-x)$ ?

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How do you solve the equation ${0.16}^{4 + 3 x} = {0.3}^{8 - x}$ $0.16^(4+3x)=0.3^(8-x)$ ?

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Answer 1 · 2017-01-29T21:51:22

$x = 0.29$ $x=0.29$

Explanation:

${0.16}^{4 + 3 x} = {0.3}^{8 - x}$ $0.16^(4+3x)=0.3^(8-x)$

Take the log of both sides

${log 0.16}^{4 + 3 x} = {log 0.3}^{8 - x}$ $log0.16^(4+3x)=log0.3^(8-x)$

${log a}^{b} = b log a$ $loga^b=bloga$

$(4 + 3 x) log 0.16 = (8 - x) log 0.3$ $(4+3x)log0.16=(8-x)log0.3$

$4 + 3 x = \frac{log 0.3}{log 0.16} (8 - x)$ $4+3x=log0.3/log0.16(8-x)$

Note: from this point on, it is better to use the exact value of $\frac{log 0.3}{log 0.16}$ $log0.3/log0.16$ , but for the sake of simplicity, a rounded value was used.

$4 + 3 x = 5.33 - 0.66 x$ $4+3x=5.33-0.66x$

$4.66 x = 1.33$ $4.66x=1.33$

$x = \frac{1.33}{4.66} = 0.29$ $x=1.33/4.66=0.29$

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How do you solve the equation ${0.16}^{4 + 3 x} = {0.3}^{8 - x}$ $0.16^(4+3x)=0.3^(8-x)$ ?

1 Answer

Explanation:

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