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

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Answer 1 · 2018-08-03T07:59:20

The solution is $x = - 1.165$ $x=-1.165$

Solve by taking the logs on both sides of the equation

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

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

$(x + 4) ln 3 = (1 - 3 x) ln 2$ $(x+4)ln3=(1-3x)ln2$

$x ln 3 + 4 ln 3 = ln 2 - 3 x ln 2$ $xln3+4ln3=ln2-3xln2$

$x ln 3 + 3 x ln 2 = ln 2 - 4 ln 3$ $xln3+3xln2=ln2-4ln3$

$x (ln 3 + 3 ln 2) = ln 2 - 4 ln 3$ $x(ln3+3ln2)=ln2-4ln3$

$x = \frac{ln 2 - 4 ln 3}{ln 3 + 3 ln 2}$ $x=(ln2-4ln3)/(ln3+3ln2)$

$x = - \frac{3.701}{3.178} = - 1.165$ $x=-3.701/3.178=-1.165$

How do you solve 3^(x+4)=2^(1-3x)3x+4=21−3x3^(x+4)=2^(1-3x)?