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.

## e^x-30e^-x=1

# 2 Answers

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,

y^{2} - 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

## Comments

Good answer, Kevin.

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