Solve each equation. Give solutions in exact form. See Example...

Question

Solve each equation. Give solutions in exact form. See Examples 5–9. log₂ (2x - 3) + log₂ (x + 1) = 1

Accepted Answer

Hello everybody, everything. All right. Today we're gonna be looking at this question that's asking us to solve the logarithmic equation for X and provide an exact solution where our equation is log of base three of three, X minus five plus log of base three of X plus four equal to one. Now, the answer choices of the providers are a negative seven plus five, multiplying the scored of 13 and all of that divided by six B, the same as A except you're subtracting an enumerator instead of adding C is both A and B and D is negative five plus a multiplying the scored of 11 and all of that divided by three. Now we're looking at the equation that we have I, the first thing I noticed is that we have two logs of the same base, adding each other on the left hand side. So with that I can do some recall on different rules that we have for algorithms. In this case, the product rule. So I'll write a general formula for that. Let's say we have log of base B of A plus log of base being of scene that will be equal to log of base B of a multiplying C. So comparing that to the equation that we have, the base three will be the base B in a recall, three, X minus five will be A in our recon and the term X plus four will be the C in a recall. So with that, let's rewrite the equation. So we'll have log of base three of three X minus five, multiplying X plus four and they'll continue to be equal to one. Now, we can do some distribution on what it's multiplying the three X minus five, multiplying the X plus four, we can distribute that. So let's do that. We have log base three of and as we distribute it'll be three X multiplying X which will be three X squared, then three X multiplying positive form which will be two plus 12 X in the equation. Then negative five, multiplying positive X which will be minus five X in the equation and finally negative five multiplying positive four which will be minus 20 in our equation and I continue to be equal to one. Now, we can do some quick simplification here when we know 12 X minus five X is gonna be seven X. So let's rewrite that to reflect that we'll have log base three of three X squared plus seven X minus 20 equal to one. Now, we can look at this as just a log of something equal to a number and by just looking at it, we can think about how we can work with this. Well, what we can do is turn this into an exponential equation. And I'll go back to our recall and write another rule that we have on how to turn logarithmic equations into exponential equations. So again, a general formula would just be, let's say we have log of base B of A equal to C that can be rewritten as B to the power C equal to A. Now, comparing that to the equation that we have, our base three will be the B in our recall, the entirety of the term three X squared plus seven X minus 20 will be the A in a recall and the answer one will be the C in our recall. So using that information, let's rewrite what we have. So we'll have three to the first power equal to three X squared plus seven X minus 20. And a simple simplification, we know three to the first power is just gonna be three. And now we can move that three over from the left hand side to the right hand side to have zero is equal to three X squared plus seven X and first we had minus 20 but now we're taking another three away. So it'll be minus 23. Now this, we can try to factor it, but we can also put it into our quadratic formula which uh I'll write it down here. So we can remember it is X is equal to negative B plus or minus the square root of B squared minus four, multiplying a multiplying C and all of that divided by two multiplying A. Now, in this case, the A, the B and the C do have a corresponding term in our equation. So the A is always gonna be associated with whatever number is in front of the X squared. So in this case, we have three X squared, so R A will be equal to three because there's a three in front of the X squared, the B is always associated with the term in front of the X. In this case, we have seven X positive seven X. So R B will be seven and the C is always associated with the number by itself. No AX is next to it. In this case, we have minus 23 at the end of our equation. So we have C equal to negative 23. Now we can plug all those values into our, our quadratic formula. So we'll have X is equal to negative B which will be negative seven plus or minus the square root of B squared. So we'll have seven squared minus four multiplying A which in this case is three multiplying C which in our case is negative 23. And then all of that will be divided by two multiplied by A which in our case is three. So some simp simplification, we know seven squared is gonna be 49 2 multiplied by three is gonna be six. So we would just distribute everything uh subtract the 49 or, or subtract four multiplied by three, multiplied by negative 23 from 49 and simplify all that just to give us two different X values. One of which is gonna be X is equal to negative seven plus five, multiplying the square root of 13 of that divided by six and X is equal to negative seven minus five, multiplying the square root of 13 with all of that divided by six as well. Now the first instinct here would be to go to answer choice C which has both of the axes that we obtained. However, before we do that, we have to test if the axes work. So in this case, if we plug in the negative seven minus five, multiplying the score to 13 divided by six, if we plug that into our equation, it won't work necessarily. So let's say you plug it into one of the logs. Let's say we plug it into log base three of three, X three, multiple. Yeah, three X minus five. Let's say you plug it into there will have a negative value. So we can't take the log of a negative value that will be inconclusive. Therefore, this X would not work. So our actual answer will be the other one because it will work we won't have a log of a negative value. Therefore, we are left with answer choice a in this scenario. So I hope that was helpful. And on the next end.

Solve each equation. Give solutions in exact form. See Examples 5–9. log₂ (2x - 3) + log₂ (x + 1) = 1

Key Concepts

Properties of Logarithms

Solving Logarithmic Equations

Domain Restrictions of Logarithmic Functions

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Solve each equation. Give solutions in exact form. See Examples 5–9. log2 (2x - 3) + log2 (x + 1) = 1

Key Concepts

Properties of Logarithms

Solving Logarithmic Equations

Domain Restrictions of Logarithmic Functions

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Solve each equation. Give solutions in exact form. See Examples 5–9. log₂ (2x - 3) + log₂ (x + 1) = 1