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. log₂(x+2)−log₂(x−5)=3

Accepted Answer

Hey everyone, welcome back in this video. We're asked to solve a logarithmic equation. We're told to reject any X value that would make the original function undefined. And to approximate our solution to three decimal places. The equation we're given is log base three of x minus six minus log base three of x plus four is equal to one. So rewriting what we're given log base three of X plus six minus log base three of x plus four equals one. Okay, when we have a problem like this, what we want to do is try to combine the logs into one single log. Okay we have the same base here. So that makes it easier. Let's go ahead and recall our log rule that deals with subtraction. If we have log base B of some value A minus log base B of some value K. We know we can write that equals to log base B of a divided by K. Okay so the subtraction of the separate quantities becomes a division. Alright so let's call this the red rule. We're going to use that rule to get to the next line. So this is going to become log base three of x plus six divided by X plus four. Okay and on the right hand side we have equals one stuff. Alright so now we have one single log and a number on the other side. Well let's recall our other log rule may we know if we have log base B of some quantity X is equal to a number A. Okay that we can rewrite this as B to the exponent A. Equals X. Okay so the base B of the log rhythm becomes the base B. Of the exponent. Using this blue rule we can write the base of the log rhythm. Three becomes the base of the exponent. So three to the exponent one. And everything within the law algorithm goes to the right hand side. X plus six. Over X plus four. Okay. All right now we've gotten rid of our log rhythm. The only thing we have left to do is to solve this equation. Ok this is just a linear equation so we know how to do that. Okay so multiplying by X plus four we get three times X plus four Is equal to X-plus six. Okay, expanding on the left, distributing three to both terms three X plus 12 is equal to X plus six. Okay, we're going to move the X to the left hand side. We'll get two. X plus 12 is equal to six. Subtracting 12 from both sides to X is equal to minus six. And finally dividing we get X is equal to minus three. Okay, so that's our potential solution and now we just need to double check that the original function is defined there. Um and that we don't need to reject that value. Okay so when we're thinking about the domain there are two terms to consider. We need to consider log base three of X plus six. Okay when we look at its domain we have X plus six and we need this to be bigger than zero. We can't take the log rhythm of zero or anything below. So we need whatever is in the log rhythm bigger than zero. That means that X needs to be bigger than minus six. Okay. Well we have minus three. That's bigger than minus six. So we get a check mark. We're okay there. Okay. The second term we need to consider is log base three of X plus four from the original function. Okay. And when we're thinking about the domain, similarly we need whatever's inside the log rhythm bigger than zero. So in this case we need X plus four bigger than zero and that's going to give us X bigger than minus four. Okay and again our value is minus three. That's bigger than minus four. So we get a check mark. Our solution is indeed a solution. Okay, so x equals -3 is our solution that is answered. D. I hope this video helped everyone. Thanks for watching

Key Concepts

Properties of Logarithms

Domain of Logarithmic Functions

Converting Logarithmic Equations to Exponential Form

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