Solve each logarithmic equation in Exercises 49–92. Be sure to...

Question

Solve each logarithmic equation in Exercises 49–92. Be sure to reject any value of x that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. ln(x−4)+ln(x+1)=ln(x−8)

Accepted Answer

Welcome back. I am so glad you're here. We are given this log rhythmic equation, we're told that the natural log of x minus three plus the natural log of X plus two equals the natural log of x minus six. And we need to solve for X. Well there are three Xs and they're all inside logs. How do we solve for X? Well, let's recall from previous lessons on expanding and condensing logs that when we have two logs that are being added to each other, then what's inside the parentheses here can be rewritten as being multiplied by each other. So this would be rewritten as the natural log of the two things inside parentheses is being multiplied by each other. So the X minus three times X plus two. And that's still equal to the natural log of X minus six. But this is a lot more helpful because now we recall Previous lessons on the 1-1 property of logarithms. And we've got a log of something on the left equals a log of something on the right. And because of that property, we know that the something on the left is equal to the something on the right and we can just rewrite it with just something equaling the something or just the X minus three times X plus two equaling the X minus six. Now the logs are gone and we can solve for X let's foil on the left hand side. So we've got X times X for our first gives us X squared X, times two. For our Outsides gives us plus two, X are inside negative three times X is minus three X in our last negative three times two is minus six equal to x minus six. And now we'll combine these two leg terms in the middle on the left hand side. So bring down X squared two, X minus three, X is minus X minus six equals x minus six. And now since we've got an X squared, this could be a quadratic. And let's bring everything on the right hand side over to the left so that we can set it equal to zero and hopefully do some factoring. So we're subtracting X from both sides and adding six to both sides on the left. Bring down our X squared negative X minus X is minus two. X. The sixes cancel out. So it's X squared minus two. X equals zero. Well, we can factor out an X. Because both of those terms on the left have an X. So it's X times and dividing both terms by X leaves us with x minus two equal zero. Well, both of those X and x minus two are factors. So we can set both of those equal to zero. We have X equals zero and we have x minus two equals zero adding to to both sides of this equation on the right will give us an X equals two. So we have two solutions. Great. This is wonderful. Except it's very important to remember that logs have domain restrictions. What's inside of the log parentheses here needs to be positive. So we need to have x minus three. X plus two and x minus six. All be positive values. They need to be greater than zero. So let's plug X equals zero in for our X value to see if they're positive. So zero minus +30 plus two and zero minus six. Okay, no zero minus three is negative. Zero minus six is negative. So X equals zero. That doesn't work. That would be undefined for our logs and we can't have that as a solution. Now let's plug two in for X value. We have two minus three to plus two and two minus six. Again to minus three. That's negative. Two minus six. That's negative. We can't have X equal to its a domain restriction. So actually after doing all of that work, the answer that matches what we found is d no solution. Well done. We'll catch you on the next one

Key Concepts

Properties of Logarithms

Domain of Logarithmic Functions

Solving Logarithmic Equations

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