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

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Answer 1 · 2015-12-01T01:24:20

$x = -(4ln(3))/(3ln(4) - ln(3))$

For this problem, we will be using the property of logarithms that

$ln(a^x) = xln(a)$

$4^(3x) = 3^(x-4)$

$=> ln(4^(3x)) = ln(3^(x-4))$

#=> 3xln(4) = (x-4)ln(3)

$=> 3xln(4) - xln(3) = -4ln(3)$

$=> x(3ln(4) - ln(3)) = -4ln(3)$

$=> x = -(4ln(3))/(3ln(4) - ln(3))$

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