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

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. log3(x+6)+log3(x+4)=1

Accepted Answer

everyone in this problem. We are asked to solve a logarithmic equation. We're told to reject any X values that make the original function undefined. And to approximate our solution to three decimal places. Now the equation we're given is log base five of x plus five plus log base five of x plus two is equal to two. Okay so rewriting what we're given log base five of X plus five plus log base five of x plus one equals two. Okay now problem like this we want to try to combine the logarithms into one. So let's recall the rule we have when we have logarithms being added together. So if we have log base B. Of something A plus log base B of something K. Well they have the same base so we can combine them and it's gonna be log base B of A. K. So this is the rule we have with addition using that rule we can write Our equation as log base five of x plus five. All times X plus one. Okay so we've combined those by multiplying that's gonna equal to. Okay now we have a single algorithm, we can go ahead and convert that using another log roll. Okay so we have log base B. Of X equals the number A. We can write that as B to the A equals X. Okay those are equivalent expressions. So the base B of the log becomes a base B of the exponent. So applying this to our Equation We're gonna have the base B squared is equal to X plus five times X plus one. And that squared comes from this two over here. Alright well now this is just a quadratic equation, we know how to solve quadratic equations, we've gotten rid of the logs so let's go ahead and do that. So on the left hand side we have 25 on the right hand side we can expand this to get X squared plus six X plus five. Moving the 25 over we have zero equals x squared plus six. X minus 20. Alright so we have a quadratic equation. We don't see any clear way to factor it. So let's go ahead and use our quadratic equation. So X is equal to its gonna be minus B plus or minus the square root of b squared minus four. A. C. All over two. Okay. And let's not get confused. We have been our quadratic equation. That's not the same. B. When we're referring to the base of the algorithm. Okay so in this case A is going to be one, B is gonna be six and C is going to be -20, Plugging in those values, we get negative six plus or minus the square root of six squared - Times one times -20. All over two. Well that's gonna give us minus six plus or minus. We're going to have 36 plus 80. Okay that's gonna be the square root All over to I'm gonna continue writing on the right here, that's -6 over two plus or minus while we can separate 100 and 16. Okay. We know that we can divide by four. That's going to give us the square root of four times all over to. Well the square root of four is two. So we're gonna get minus three plus or minus two times the square root of 29. All divided by two. Okay so we've pulled that four out. Taking the square root, the twos will cancel and we're gonna be left with minus three plus or minus the square root of 29. Okay I'm just gonna scroll down. We have a little bit more room to right here. All right so minus three plus or minus square root of 29. And if we use our calculator in round 23 decimal places like we were asked to in the question we get minus 8. or 2.385. Okay so those are two potential solutions. Now what we need to do is we need to look at do those satisfy the original equation. Now there's two places we need to look in the original equation. Okay so we need to consider the term log base five of X plus five. Okay and when we're looking at this we know that whatever we have inside of the log needs to be bigger than zero. We can't take the log of zero or anything less. So we need X plus five bigger than zero. Okay. And that's gonna tell us that we need X bigger than -5. Alright. So when we're looking at the solutions we have 2.385. That's bigger than minus five but minus 8.385. That's not bigger than minus five. Okay So if we use that in the original equation we're going to have an undefined equation so that one is not a solution. Okay. Now the second part that we need to check on is log base five of X plus one. Okay. And again thinking about the domain we need what's inside the log to be bigger than zero. That's gonna tell us that X has to be bigger than minus one. Ok. And in this case the other solution we have 2.385. That is still bigger than minus one. And so that one is indeed a solution. Okay, so X equals 2.385 is our solution that's going to correspond with answer A. I hope that helped. Thanks for watching everyone see you in the next video

Key Concepts

Logarithmic Properties

Domain of Logarithmic Functions

Solving Logarithmic Equations

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