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# C2 - Logarithms

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1. 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

However, the textbook says the answer is 2?

42x = 100
2. (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:

logx(2x) = 2

However, the textbook says the answer is 2?

42x = 100

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.
3. (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:

logx(2x) = 2

However, the textbook says the answer is 2?

42x = 100
how can x = sqrt2 x?
that would mean x = 0

and you can't take log base 0
4. (Original post by SeanFM)

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.

For the second question, I done two methods:

[latex]4^2x = 1000
log4(1000) = 2x
2x = 4.98
x = 2.49

4^2x = 1000
log10(4^2x) = 1000
2x log10 4 = 1000
2x = 1000 / log10 4
x = ans above / 2
5. (Original post by Student403)
how can x = sqrt2 x?
that would mean x = 0

and you can't take log base 0
Idk, I just started logarithms today or fun. I don't know what I'm doing, so I need help.
6. (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
log4(1000) = 2x
2x = 4.98
x = 2.49

4^2x = 1000
log10(4^2x) = 1000
2x log10 4 = 1000
2x = 1000 / log10 4
x = ans above / 2
you said in the OP that 4^2x = 100, and you're working it out as 1000
7. (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
log4(1000) = 2x
2x = 4.98
x = 2.49

4^2x = 1000
log10(4^2x) = 1000
2x log10 4 = 1000
2x = 1000 / log10 4
x = ans above / 2
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.

8. Q4:
logx(2x)=2
x2=2x
x2-2x=0
x(x-2)=0 therefore x=0 or x=2 ; ignore x=0 as log(0) is undefined for any base.

Q1:
42x = 100
2x=log4(100)
x=log4(100) / 2
x is approximately 1.66
9. (Original post by Student403)
you said in the OP that 4^2x = 100, and you're working it out as 1000
Sorry, thanks! Dam, such a stupid mistake.

(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.

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
To get x in brackets, you square root 2x.
10. (Original post by RDKGames)
Q4:
logx(2x)=2
x2=2x
x2-2x=0
x(x-2)=0 therefore x=0 or x=2 ; ignore x=0 as log(0) is undefined for any base
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.
11. (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 cross-check your answer in 2 ways lol.

For the first part, what I'm saying is:
logx(2x) = 2
So, that means
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.

using various properties of logs.
12. (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.
You're welcome! Well you were technically RIGHT, but it doesn't help finding the values of x for which it is true xD
13. (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.
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.
14. (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.
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).
15. (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.
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:
16. (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).
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.

(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 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.
17. (Original post by RDKGames)
x
(Original post by SeanFM)
x
(Original post by Student403)
x

(g) 32x+1 - 26(3x) - 9 = 0
(h) 4(32x+1) + 17(3x) - 7 = 0

How I solved them:

(g)

log39 = x
x = 2

When y = -1/3, there are no solutions.

(h)

log31/3 = x
x = -1

When y = -7/4, there are no solutions.
18. (Original post by Chittesh14)

(g)
(h)
Is that latex being annoying when it should be ?
19. (Original post by SeanFM)
Is that latex being annoying when it should be ?
Yeah lol, sorry I'm not too good with latex.
I've wrote them again.
20. (Original post by Chittesh14)

(g) 32x+1 - 26(3x) - 9 = 0
(h) 4(32x+1) + 17(3x) - 7 = 0
Right.. and how have you solved them / why do you think its taking so much time? It could be that you're just getting used to it at the start.

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