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e^x-30e^-x=1

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2 Answers

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

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