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

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Answer 1 · 2015-12-01T01:24:20

$x = - \frac{4 ln (3)}{3 ln (4) - ln (3)}$ $x = -(4ln(3))/(3ln(4) - ln(3))$

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

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

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

$\Rightarrow ln (4^{3 x}) = ln (3^{x - 4})$ $=> ln(4^(3x)) = ln(3^(x-4))$

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

$\Rightarrow 3 x ln (4) - x ln (3) = - 4 ln (3)$ $=> 3xln(4) - xln(3) = -4ln(3)$

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

$\Rightarrow x = - \frac{4 ln (3)}{3 ln (4) - ln (3)}$ $=> x = -(4ln(3))/(3ln(4) - ln(3))$

How do you solve 43x=3x−44^(3x)=3^(x-4)?