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Algebraically (not graphically) solve the following equation To receive full credit, you must showyour work:4.8 = loga.2 (62 + 2)...

Question

Algebraically (not graphically) solve the following equation To receive full credit, you must showyour work:4.8 = loga.2 (62 + 2)

Algebraically (not graphically) solve the following equation To receive full credit, you must showyour work: 4.8 = loga.2 (62 + 2)



Answers

Solve the equation both algebraically and graphically. $$ 6(x+2)^{5}=64 $$

So here we have an equation. Four over Express to buy a 6/2 X equals 5/2 X Plus four and were asked to sell this equation. Now the first day I'm going to do is make sure that everybody has the same denominator because that makes it easier to work with. So this tsunami can be written as to Times X plus two. So this case, the least common denominated would be two X Times X plus two. So I will multiply each fraction by an equivalent fraction. That's equal to one. But we'll get the denominated two x Times X plus two. So this one we will have to multiply by two x over two x. This faction we will have to multiply by X plus two over X Plus two, and this faction is kind of messy. But we will have to multiply by X over X so I will get a a X over two x Times X plus two minus six uh minus 12 over two X hives explicit to equals five x over two X Times X plus two. So now that they all the same denominator, I can simply add the I'm going to remove this crap down a little bit because we don't need it yet. Sorry, that's like in the way we'll just push it off to the side. So now we have a x My six x is two x two X minus 12 Over two x times X plus two equals five x over two X Times expressed All express to So I Did I say 12 limit to Okay, No, let's multiply both sides by two x Times X plus two. So we get out of the denominator and then it will cancel out both the top in the bottom of both sides. So basically goes away. To know we have simply to x minus 12 equals five X. Now let's subtract two x from both sides. So we get negative. 12 equals three x divide both sides by three and we get X equals negative before as our solution to that equation. Now that's how we found the solution. Algebraic Lee. Let's see if we can find the same solution, but graphically so. What I would do is I would grow four over x plus two by a 6/2 x as one function. And then I would grab 5/2. X Plus four has another function. So we would take those two functions and we would put them into a graphing calculator. And we'll end up with something that looks like this. And now we need to find where these two graphs and dissect each other. And if yes, where they intersect, right? If years of the Intersect right here and that is ah, X equals negative for with. So that is how we found the solution. Graphically. And as you can see, it is the same as the solution we found, Algebraic Lee.

To solve this equation. I would first get everything on the same side by subtracting the six. And you could if you wanted to look at it like this with positive exponents, because we're gonna try to factor. That might make your factoring a little bit easier to see. So the three can be a three and a one. The why did the fourth is a Y squared in a Y squared. So the two is a two in a one, and I need to get a six in the middle. Mm. And I don't think that is possible. So I am not going to factor. I'm gonna use the quadratic formula instead. Okay, So are a value is three R b value negative six and C is to so the opposite of B plus or minus the square root of B squared minus for a see all over two A. So we've got six plus two minus 36 minus 24 over six, six plus or minus the square root of 12/6. Now, don't be tempted to cancel those six outs. Six sixes out because you can't. So I'm gonna do six. Plus the square were to 12/6 and six, minus the squirt of 12/6 and get two answers. So this one is, what, 1.577 this 1.4 to. Okay, so then let's go to our graphing calculator and wrong picture. We could put the left side into y one and the right side into y two, and then we can grab that. And you want to calculate the intersection by doing a second trace Number five. And let's see the first intersection. I'm going to take a picture of this. Mhm. Okay, so here is my first intersection point. So you could see there's the one answer right there after it's pretty close. And then I'm gonna have toe calculate another intersection like you do have to move the cursor over by your intersection on to be enjoying my Halloween tablecloth. Okay. And then this one positive 1.5 Mhm. Let me go back. So I got the positive. I also got a 0.42 which is not my other answer that I got Okay. The the reason why I didn't get those two answers is I did forget that race that line this waas a ah y to the whips. I was on a race a y to the fourth problem. So really, I should have, like, a plus or minus in front of both of these plus or minus plus or minus. So my graph gave me these two answers, and I really did not see these two. So I'm thinking if we go back to the beginning like those in tow, our equation, they'll give us an error. We won't be able to get those answers unless there that unless the intersection is so my Newton happens at two places that I did not see. Try this again. I will try to zoom in on my picture a little bit more, and I want to see maybe if it does cross over twice. Took the bear with me while I graft this. Oh, it does. Okay, I'm gonna get this picture a little bit better for you, and I'm gonna show you that graph. I could show you yet you do get all four answers. Okay, Bring my It's still graphing here, but there you go. And I'm gonna take a picture of this. So you've really had to get in close to this graph to see Yes, that there were all for the answers. Okay, because I zoomed into right in here on this one to see that a little bit better. So you can see there are two. So this is the negative 1.5, and then this is the negative 0.4 to the same thing happens over at the positive side of the graph because it's, um, symmetric, but the Y axis.

Altra Brackley. We would solve this by dividing by six and then taking the fifth fruit and then subtracting to isolate X Graphically, we're going to solve this by inserting each side of the equation as its own Grax. So we will have y equals six times X plus two do the fifth power and then we will also have the graph of y equals 64. What this does is it allows us to look for areas of intersection. Ah, this point of intersection will tell us that this is a solution of the equation. So here we see that the intersection occurs when X is equal to negative 0.395 So that means that X equals negative. 0.395 is a solution to the equation.

Okay, so here we have any question off parabola and we want to solve this equation for X using this graph. So first of all, divided by three, both side and 48 divided by 3. 16. Now subtract 16 both side. Now, to grab dysfunction, we need a few points. So make a table values. And here why is equal to the old function that is experiments 16. Okay, s So when X is equal to minus four, then why is zero when X is a called of minus two? Then why it's minus 12 when x zero, Why is minus 16 and when X's to y's minus throat when excess four and wise zero now draw a parabola which is passing through all these five points. And from tha graph, we can see that our X intercept, our negative four and four. Therefore, our solution is X is equal to minus four and X is equal to four. But we have to check our solution and typically also. So our given equation is three. X square is equal to 48 substitute divorced value of facts which is minus four. Okay, so here, minus Warhol square, that is 16 and 16 multiplied by three is 48. So here are both sides are equal. Therefore, our solution X they call two minus four is correct. Now substitute the other value of facts which is four. So you're three multiplied by four. Ball square is the Goto 48 four square is 16 and 16 Multiply batteries 48. So again here both sides are equal. Therefore, our solution X difficult forests Correct. Thank you.


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