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:
log_{x}(2x) = 2
I wrote the answer as
However, the textbook says the answer is 2?
Question: 1 (d) Solve, giving your answers to 3 s.f.:
4^{2x} = 100
My answer: 2.49 and textbook answer: 1.66
C2  Logarithms
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 23062016 00:29

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 23062016 00:34
(Original post by Chittesh14)
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:
log_{x}(2x) = 2
I wrote the answer as
However, the textbook says the answer is 2?
Question: 1 (d) Solve, giving your answers to 3 s.f.:
4^{2x} = 100
My answer: 2.49 and textbook answer: 1.66
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. 
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 3
 23062016 00:37
(Original post by Chittesh14)
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:
log_{x}(2x) = 2
I wrote the answer as
However, the textbook says the answer is 2?
Question: 1 (d) Solve, giving your answers to 3 s.f.:
4^{2x} = 100
My answer: 2.49 and textbook answer: 1.66
that would mean x = 0
and you can't take log base 0 
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 4
 23062016 00:38
(Original post by SeanFM)
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.
Through this, i got square root 2x.
For the second question, I done two methods:
[latex]4^2x = 1000
log_{4}(1000) = 2x
2x = 4.98
x = 2.49
4^2x = 1000
log_{10}(4^2x) = 1000
2x log_{10} 4 = 1000
2x = 1000 / log_{10} 4
x = ans above / 2 
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 23062016 00:40
(Original post by Student403)
how can x = sqrt2 x?
that would mean x = 0
and you can't take log base 0 
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 23062016 00:41
(Original post by Chittesh14)
For the first question, I just worked out the square root of 2x.
Through this, i got square root 2x.
For the second question, I done two methods:
[latex]4^2x = 1000
log_{4}(1000) = 2x
2x = 4.98
x = 2.49
4^2x = 1000
log_{10}(4^2x) = 1000
2x log_{10} 4 = 1000
2x = 1000 / log_{10} 4
x = ans above / 2 
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 7
 23062016 00:42
(Original post by Chittesh14)
For the first question, I just worked out the square root of 2x.
Through this, i got square root 2x.
For the second question, I done two methods:
[latex]4^2x = 1000
log_{4}(1000) = 2x
2x = 4.98
x = 2.49
4^2x = 1000
log_{10}(4^2x) = 1000
2x log_{10} 4 = 1000
2x = 1000 / log_{10} 4
x = ans above / 2
For your first q, I'm not sure what your working meansLast edited by SeanFM; 23062016 at 00:43. 
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 23062016 00:47
Q4:
log_{x}(2x)=2
x^{2}=2x
x^{2}2x=0
x(x2)=0 therefore x=0 or x=2 ; ignore x=0 as log(0) is undefined for any base.
Q1:
4^{2x} = 100
2x=log_{4}(100)
x=log_{4}(100) / 2
x is approximately 1.66Post rating:2 
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 23062016 00:48
(Original post by Student403)
you said in the OP that 4^2x = 100, and you're working it out as 1000
(Original post by SeanFM)
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
For the first part, what I'm saying is:
log_{x}(2x) = 2
So, that means
To get x in brackets, you square root 2x. 
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 23062016 00:50
(Original post by RDKGames)
Q4:
log_{x}(2x)=2
x^{2}=2x
x^{2}2x=0
x(x2)=0 therefore x=0 or x=2 ; ignore x=0 as log(0) is undefined for any base 
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 11
 23062016 00:51
(Original post by Chittesh14)
Sorry, thanks! Dam, such a stupid mistake.
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 crosscheck your answer in 2 ways lol.
For the first part, what I'm saying is:
log_{x}(2x) = 2
So, that means
To get x in brackets, you square root 2x.
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.
using various properties of logs.Last edited by SeanFM; 23062016 at 00:53. 
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 23062016 00:52
(Original post by Chittesh14)
Wow thanks, can't believe I went from x^2 = 2x to x = square root 2x lol. Now, I understand where I blundered, so just factorise it. 
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 23062016 00:53
(Original post by SeanFM)
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. 
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 14
 23062016 00:55
(Original post by Chittesh14)
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.
See the edit in my post if you're interested in a more roundabout way of doing questions (that uses different properties of logs). 
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 23062016 00:57
(Original post by Chittesh14)
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. 
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 23062016 16:06
(Original post by SeanFM)
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).
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.
(Original post by RDKGames)
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 log_{a}(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: 
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 23062016 16:38
(Original post by RDKGames)
x(Original post by SeanFM)
x(Original post by Student403)
x
(g) 3^{2x+1}  26(3^{x})  9 = 0
(h) 4(3^{2x+1}) + 17(3^{x})  7 = 0
How I solved them:
(g)
log_{3}9 = x
x = 2
When y = 1/3, there are no solutions.
(h)
log_{3}1/3 = x
x = 1
When y = 7/4, there are no solutions.Last edited by Chittesh14; 23062016 at 16:54. 
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 23062016 16:39

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 23062016 16:41
I've wrote them again. 
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 23062016 16:44
(Original post by Chittesh14)
Questions: Solve, giving your answer to 3 s.f.
(g) 3^{2x+1}  26(3^{x})  9 = 0
(h) 4(3^{2x+1}) + 17(3^{x})  7 = 0
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Updated: July 21, 2016
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