How do you solve ln2-ln(3x+2)=1?

Jul 13, 2016

x = (2(1/e-1))/3 ≈ -0.42141

Explanation:

In order to solve this logarithmic equation, we can make use of the properties of logarithms, such as

Property:
$\ln \left(a\right) - \ln \left(b\right) = \ln \left(\frac{a}{b}\right)$

We can now rewrite this equation as follows:

$\ln \left(\frac{2}{3 x + 2}\right) = 1$

To get rid of the natural logarithm on the left-hand side, we take the $e$-xponential on both sides, giving us

$\frac{2}{3 x + 2} = {e}^{1}$

To simplify this even further and solve for $x$, the best thing to do here would be to to get rid of the fraction. We can do this by multiplying both sides by $3 x + 2$, which yields

$\frac{2}{\cancel{\left(3 x + 2\right)}} \cdot \cancel{\left(3 x + 2\right)} = e \left(3 x + 2\right)$

So our equation now has become a lot more appealing:

$2 = e \left(3 x + 2\right)$

Since $e$ is just a constant, namely e ≈ 2.718, we can divide both sides $e$ to isolate the term with the $x$ in it.

$\frac{2}{e} = 3 x + 2$

Subtracting $2$ from both sides gives us

$\frac{2}{e} - 2 = 3 x$

Dividing then by $3$ yields

$\frac{\frac{2}{e} - 2}{3} = x$

Or we can factor out a $2$ and write:

x = (2(1/e-1))/3 ≈ -0.42141