In Exercises 93–102, solve each equation. ln 3−ln(x+5)−ln x=0

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Hello today we're going to be solving for X in the given equation. So the equation is the natural log of two minus the natural log of X plus one minus the natural log of two. X is equal to zero. Now one thing that you can go ahead and do is go ahead and add the value of the natural log of two X to both sides of the equation. And doing so is going to give us the natural log of two minus the natural log of X plus one is equal to the natural log of two X. Now on the left hand side we can simplify this using the rule of subtraction for natural logs. And that states that if you have the natural log of a minus the natural log of B. That is equal to the natural log of a over B. So applying this rule to the right hand left hand side is going to give us the natural log of two over X plus one is equal to the natural log of two X. Now, since we have a natural log on both sides of the equation we can just go ahead and equate the inside arguments to each other. So doing so is going to give us two over X plus one is equal to two X. And on the right hand side we can rewrite two X as two X over one. And to simplify this, all we need to do is to go ahead and cross multiply doing so is going to give us two X times the quantity X plus one is equal to two times one which is two. And on the left hand side we're going to distribute the value of two X into the quantity X plus one. This is going to give us two, X squared plus two X is equal to two. Now, since we want to solve for X we're going to subtract two to both sides of the equation to set the equation equal to zero and this is going to give us two. X squared plus two. X minus two is equal to zero. Now if you notice on the left hand side, all the terms have the common factor of two. So if we divide this entire equation by two, what we are left with is x squared plus x minus one is equal to zero. Now this unfortunately is not a fact. Herbal trin Amiel. However, we can solve this using the quadratic equation. So first let's go ahead and label our coefficients A is going to equal to one. B is equal to one and C is equal to negative one. And we're going to apply this to when X is equal to negative B plus or minus the square root of B squared minus four times a times C. All over two times A. So plugging in our coefficients is going to give us X is equal to negative B which is negative one plus or minus the square root of b squared which is one squared minus four times A which is one times C which is negative one, All over two times A which is two times one Now X simplifying this one squared is going to be one and negative four times one times negative one is going to give us positive four. So X is equal to negative one plus or minus the square root of one plus four. All over two times one which is two and one plus four is going to give us five so X is equal to negative one plus or minus the square root of five. All over two. Now this is going to give us two values for X. X is equal to negative one plus the square root of 5/2 and X is equal to negative one minus the square root of five over to. Now since we are working with the natural logs we need to go ahead and verify the given values and that means that we're going to go ahead and plug in our values for X back into the original equation. So let's go ahead and start with the positive value of X and we can go ahead and work with the decimal value of X. So negative one plus the square root of 5/2 is going to give us X is equal to 0.618. And if we plug this into the original equation we have the natural log of two minus the natural log of 0. plus one minus the natural log of two times 0.618 is equal to zero. And if you simplify these natural logs, What you're gonna be left with is zero is equal to zero which is a true statement. So that means that X is equal to 0.618 is going to be the solution for X. But now we need to verify the negative value, so X is equal to negative one minus the square root of 5/2 has a rough estimate value of X is equal to negative 1.618. And if we plug this into the original equation we get the natural log of two minus the natural log of negative 1.618 plus one minus the natural log of two times negative 1.618 is equal to zero. Now if you notice one thing negative one plus plus one is going to give us a negative value inside the natural log and two times negative 1.618 is going to give us a negative value in the natural log. So on the left hand side we have two values that are going to be undefined And that means that X is equal to negative 1.618 is not going to be a solution for X. So there's only one solution for x X is equal to negative one plus the square root of 5/2 or in decimal form, X is equal to 0.618. And that means that the solution to this problem is going to be C. So I hope this video helps you in understanding how to solve this type of equation, and I'll go ahead and see you all in the next video.

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In Exercises 93–102, solve each equation. ln 3−ln(x+5)−ln x=0

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