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

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Answer 1 · 2017-01-29T21:51:22

$x=0.29$

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

Take the log of both sides

$log0.16^(4+3x)=log0.3^(8-x)$

$loga^b=bloga$

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

$4+3x=log0.3/log0.16(8-x)$

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

$4+3x=5.33-0.66x$

$4.66x=1.33$

$x=1.33/4.66=0.29$

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