Hey guys, I done a few questions on logarithm just now and I feel that the textbook answers are wrong, can someone guide me here or tell me if the textbook answers are right (and I'm wrong) or vice versa lol?
Question: 4 (d) Find the value for x for which:
logx(2x) = 2
I wrote the answer as 2x However, the textbook says the answer is 2?
Question: 1 (d) Solve, giving your answers to 3 s.f.:
42x = 100 My answer: 2.49 and textbook answer: 1.66
Hey guys, I done a few questions on logarithm just now and I feel that the textbook answers are wrong, can someone guide me here or tell me if the textbook answers are right (and I'm wrong) or vice versa lol?
Question: 4 (d) Find the value for x for which:
logx(2x) = 2
I wrote the answer as 2x However, the textbook says the answer is 2?
Question: 1 (d) Solve, giving your answers to 3 s.f.:
42x = 100 My answer: 2.49 and textbook answer: 1.66
Can confirm that the answer to your first question is 2 - what have you tried to get to your answer?
The second question also has the right answer already (it is worth noting that you can check them in with the original equation given) - how did you get to 2.49?
What's important here is to show your working so that you can be pointed in the right direction.
Hey guys, I done a few questions on logarithm just now and I feel that the textbook answers are wrong, can someone guide me here or tell me if the textbook answers are right (and I'm wrong) or vice versa lol?
Question: 4 (d) Find the value for x for which:
logx(2x) = 2
I wrote the answer as 2x However, the textbook says the answer is 2?
Question: 1 (d) Solve, giving your answers to 3 s.f.:
42x = 100 My answer: 2.49 and textbook answer: 1.66
Can confirm that the answer to your first question is 2 - what have you tried to get to your answer?
The second question also has the right answer already (it is worth noting that you can check them in with the original equation given) - how did you get to 2.49?
What's important here is to show your working so that you can be pointed in the right direction.
For the first question, I just worked out the square root of 2x. Through this, i got square root 2x.
Your method is correct, but for the fact that 100 has become 1000. (personally I wouldn't have bothered with log to base 4, just log (base 10 implied) both sides to give 2xlog4 = log1000, so log1000 / 2log4 = x but I suppose your method is more succint.
For your first q, I'm not sure what your working means
Your method is correct, but for the fact that 100 has become 1000. (personally I wouldn't have bothered with log to base 4, just log (base 10 implied) both sides to give 2xlog4 = log1000, so log1000 / 2log4 = x but I suppose your method is more succint.
For your first q, I'm not sure what your working means
Lol, just developed a few skills whilst learning logarithms. You can apply them to this question, so there are several methods to answering a question - even though the book teaches you 1. That's the good thing because you can cross-check your answer in 2 ways lol.
For the first part, what I'm saying is: logx(2x) = 2 So, that means (x)2=2x To get x in brackets, you square root 2x.
Lol, just developed a few skills whilst learning logarithms. You can apply them to this question, so there are several methods to answering a question - even though the book teaches you 1. That's the good thing because you can cross-check your answer in 2 ways lol.
For the first part, what I'm saying is: logx(2x) = 2 So, that means (x)2=2x To get x in brackets, you square root 2x.
I see.. interesting. It half works, but you've hit a brick wall. You can't have your answer in terms of x - you're supposed to find x.
Using what you've done (and you're right up to this point) x^2 - 2x = 0. By factorising, you get ... and one of the solutions is x = 0 but that can't work, as both log to base 0 of anything and log0 to any base is a nightmare.
I see you've worked out (been given) the answer above, so here's an alternative to that one as well.
logx2x=2⇒logx2+logxx=2⇒logx2+1=2⇒logx2=1⇒x1=2⇒x=2 using various properties of logs.
I see.. interesting. It half works, but you've hit a brick wall. You can't have your answer in terms of x - you're supposed to find x.
Using what you've done (and you're right up to this point) x^2 - 2x = 0. By factorising, you get ... and one of the solutions is x = 0 but that can't work, as both log to base 0 of anything and log0 to any base is a nightmare.
xD, thank you. I'm just doing this for fun, I'l actually be starting from Friday lol. Really fun though, found it hard lol then simple.
xD, thank you. I'm just doing this for fun, I'l actually be starting from Friday lol. Really fun though, found it hard lol then simple.
No worries. Logs are nice when you get used to them. The hardest thing after all of it is deciding whether to use x^n or x^n+1 in questions, but not to worry about that for now.
See the edit in my post if you're interested in a more roundabout way of doing questions (that uses different properties of logs).
xD, thank you. I'm just doing this for fun, I'l actually be starting from Friday lol. Really fun though, found it hard lol then simple.
I just think of logs as shifting between bases and their indices, not even sure if that makes sense. If you want a tip for logs in C2, just thing loga(b)=c as "What number, c, do I need in order to raise the base, a, by it to get b?" at least that's how I remember it. c:
No worries. Logs are nice when you get used to them. The hardest thing after all of it is deciding whether to use x^n or x^n+1 in questions, but not to worry about that for now.
See the edit in my post if you're interested in a more roundabout way of doing questions (that uses different properties of logs).
Oh yeah, I know those formulaes. I just started logs yesterday so I'm not used to them enough to be able to remember and apply them in these questions.
I just done 2 more log questions and I was wondering if there is a faster way of solving them. I will post them in around 10 minutes.
I just think of logs as shifting between bases and their indices, not even sure if that makes sense. If you want a tip for logs in C2, just thing loga(b)=c as "What number, c, do I need in order to raise the base, a, by it to get b?" at least that's how I remember it. c:
Yeah, that's how I think of it too lol. But, sometimes when they ask you to solve for a or c, it's a bit annoying when the numbers are weird. But, that's when you obviously use the calculator lol.
(g) 32x+1−26(3x)−9=0[br]32x−26(3x−1)−3=0[br]32x−26313x−3=0[br]Lety=3x[br]y2−326y−3=0[br]3y2−26y−9=0[br]3y2+y−27y−9=0[br]3y(y+1/3)−27(y+1/3)=0[br](3y−27)(y+1/3)=0[br](y−9)(3y+1)=0[br]y=9ory=−1/3[br][br]Wheny=9,3x=9 log39 = x x = 2
When y = -1/3, there are no solutions.
(h) 4(32x+1)−17(3x)−7=0[br]4(32x)+17(3x−1)−7/3=0[br]Lety=3x[br]4y2−317y−7/3=0[br]12y2−17y−7=0[br]12y2−4y+21y−7=0[br]12y(y−1/3)+21(y−1/3)=0[br](12y+21)(y−1/3)=0[br](4y+7)(3y−1)=0[br]y=−7/4ory=1/3[br][br]Wheny=1/3,3x=1/3 log31/3 = x x = -1