the "e" in this equation deals with logarithms. I understand that you change the ''e'' to ln which means natural log but then I don't know what to do next.
Sometimes a change of variables makes the problem easier. In this instance, we can let y = ex
Then we have y - 30/y = 1 or after multiplying by y, and moving everything to the left side,
y2 - y - 30 = 0
(y + 5)(y - 6) = 0
y = -5 or 6
Now the only real solution is x = ln 6.
x = ln -5 is a complex number as you will learn in more advanced math classes. ln fact ln -5 = ln 5 + πi.
the natural log does come into play, but later in the problem.
Let's start by clearing out the e-x by multiplying each side by ex:
ex(ex) - (ex)(30/ex) = ex
e2x - 30 = ex . Now let's move the ex to the other side:
e2x - ex - 30 = 0 And this looks a lot like a polynomial that needs factoring:
(ex + 5)(ex - 6) = 0
ex = -5
or ex = 6
Now the solution ex = -5 we need to discount, because e (a positive number) cannot be raised to a power to get a negative result.
So let's examine ex = 6. Now we use the ln function:
ln 6 = x . Plug ln 6 into the calculator,
x = 1.79175