Sometimes a change of variables makes the problem easier. In this instance, we can let y = e^x

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 e^x:

e^x(e^x) - (e^x)(30/e^x) = e^x

e^2x - 30 = e^x .  Now let's move the e^x to the other side:

e^2x - e^x - 30 = 0   And this looks a lot like a polynomial that needs factoring:

(e^x + 5)(e^x - 6) = 0

e^x = -5    
or e^x = 6

Now the solution e^x = -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 e^x = 6. Now we use the ln function:

ln 6 = x   . Plug ln 6 into the calculator, 

x = 1.79175