Question

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

Answer 1

`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(a) - ln(b) = ln(a/b)`

We can now rewrite this equation as follows:

`ln(2/(3x+2)) = 1`

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

`2/(3x+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 `3x+2`, which yields

`2/(cancel((3x+2)))*cancel((3x+2)) = e(3x+2)`

So our equation now has become a lot more appealing:

`2 = e(3x+2)`

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.

`2/e = 3x+2`

Subtracting `2` from both sides gives us

`2/e - 2 = 3x`

Dividing then by `3` yields

`(2/e - 2)/3 = x`

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

`x = (2(1/e-1))/3 ≈ -0.42141`