Using natural logs to solve for x

**Rae** · Dec 7th 2010, 08:57 PM

Solve for x: 3e^(-x) = 4^(3x-1)

Thanks

**harish21** · Dec 7th 2010, 09:18 PM

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

$\dfrac{3}{e^x}=4^{3x-1}$

$3 = e^{(x)}\cdot 4^{3x-1}$

take log on both sides:

$\log{(3)} = \log\bigg(e^{(x)}\cdot 4^{3x-1}\bigg)$

now use $\log(AB)=\log(A)+\log(B)$ on the RHS.

**HallsofIvy** · Dec 8th 2010, 02:31 AM

Or, more directly, take the logarithm of both sides of the original equation:
$ln(3e^{-x})= -x+ ln(3)= ln(4^{3x-1})= (3x-1)ln(4)$

Now you have -x+ ln(3)= 3ln(4)x- ln(4) which is a linear equation.

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