How do you solve $3^(x+4)=2^(1-3x)$ ?

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Answer 1 · 2018-08-03T07:59:20

The solution is $x=-1.165$

Solve by taking the logs on both sides of the equation

$3^(x+4)=2^(1-3x)$

$ln(3^(x+4))=ln(2^(1-3x))$

$(x+4)ln3=(1-3x)ln2$

$xln3+4ln3=ln2-3xln2$

$xln3+3xln2=ln2-4ln3$

$x(ln3+3ln2)=ln2-4ln3$

$x=(ln2-4ln3)/(ln3+3ln2)$

$x=-3.701/3.178=-1.165$

How do you solve 3^(x+4)=2^(1-3x)3^(x+4)=2^(1-3x)?