Solve each equation. In Exercises 11–34, give irrational solution...

Question

Solve each equation. In Exercises 11–34, give irrational solutions as decimals correct to the nearest thousandth. In Exercises 35-40, give solutions in exact form. See Examples 1–4.
e^2x - 6e^x + 8 = 0

Accepted Answer

Hello, today we're gonna be solving the following equation for X and then providing the exact solution. So what we are given is E raced to the power of two X minus 16 E to the X plus is equal to zero. Now, there's one thing that we can do. In order to approach this problem, we can go ahead and use a substitution in the given trinomial. See the first term E to the two X can be rewritten as E raced to the power of X, raced to the power of two. And so if we rewrite this first term as E race to the power of X, race to the power of two, the new trinomial we get is E raced to the power of X race to the power of two minus 16 E to the power of X plus 48 is equal to zero. Now notice that we have two instances of E to the power of X, we can go ahead and apply a substitution, let's say A and set that equal to E to the power of X. We can go ahead and simplify the current trinomial and rewrite this as a squared minus 16, A plus 48 is equal to zero. Now, let's go ahead and simplify the trinomial. So currently, a squared minus 16 A plus 48 is in the form of a factor of trinomial. And through the trial and error method, this is going to factor to give us a minus 12 times a minus four equal to zero. Now, using the zero property, we can set both factors of a equal to zero and solve for a. So doing this is going to give us a minus 12 is equal to zero and a minus four is equal to zero. On the left hand side, we're going to add 12 to both sides of the equation. And that is going to give us A is equal to 12. And on the right hand side, we're going to add four to both sides of the equation and that is going to give us A is equal to four. Now keep in mind that we want to solve for X, but currently we have E two equations in terms of A. So we're going to have to resue our original substitution. So since A is equal to E to the power of X replacing A with that value is going to give us E to the power of X is equal to 12 and E to the power of X is equal to four. So let's go ahead and solve for both of these equations. So we're going to first start with E to the power of X is equal to 12. And there is one property that we can do. In order to get the exponent of X down from the exponential value, we can set both sides of the equation as natural logarithms. So if we take the natural logarithm of the left hand side and right hand side, we can rewrite the equation as the natural log of E to the power of X equal to the natural log of 12. And there is one property for natural logarithms that we can use in order to get the X down from the exponential spot. That logarithmic property states that if you have the natural log of a race to the power of B, you can bring the exponent in front of the natural log as a product. Thus, we can rewrite the natural log of A to the power of B as B times the natural log of A. And if we apply this property to the natural log of E to the power of X, we can rewrite this equation as X times the natural log of E equal to 12. I'm sorry, not 12, the natural log of 12. And there is one more value that we need to keep in mind the natural log of E is always going to simplify, to give us one. So we can replace the natural log of E with one. And that leaves us with X times one on the left hand side of the equation, which is just going to give us X. So we get X is equal to the natural log of 12, which is going to be one of our solutions. And now we just need to solve for the other equation. So on the second equation, we have E to the power of X equal to four repeating this process, we're going to set both sides of the equation as a natural log. And we can rewrite this equation as the natural log of E to the power of X equal to the natural log of four. Now reusing the, the power property for natural logs, we can go ahead and rewrite the equation as X times the natural log of E equal to the natural log of four. And using the natural log of E is equal to one, we can go ahead and replace the natural log with E with one and that is going to leave us with X times one, which is equal to X. And so that is going to give us X is equal to the natural log of four, which is going to be the second solution to the given equation. So just to summarize, we have X is equal to the natural log of 12 and X is equal to the natural log of four as our solutions to the original equation. And with that being said, the answer to this problem is going to be a. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

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Solve each equation. In Exercises 11–34, give irrational solutions as decimals correct to the nearest thousandth. In Exercises 35-40, give solutions in exact form. See Examples 1–4. e^2x - 6e^x + 8 = 0