BBeyond
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#1
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#1
I usually get most topics in maths, but I'm really struggling with logs for some reason

Take the attached question for example: Image

I don't even know where to begin on questions like this? I know all the rules but I just don't seem to be able to apply the right ones in the right situation?

Is there anyway to fix this besides just practicing because I don't really feel like I'm getting anywhere
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Zacken
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#2
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(Original post by BBeyond)
I usually get most topics in maths, but I'm really struggling with logs for some reason

Take the attached question for example: http://gyazo.com/edf4932d8926504774c2d47b24bdb67e

I don't even know where to begin on questions like this? I know all the rules but I just don't seem to be able to apply the right ones in the right situation?

Is there anyway to fix this besides just practicing because I don't really feel like I'm getting anywhere
Well, generally, you'd want to simplify down the equations as much as you can, in this case, it would be to reduce them down to simple linear equations in x and y.

For the first equation: 8^y = 4^{2x+3}, we could reduce this down to a simple linear equation by making both sides of the equation in the form of 4^k. In this case, 8 = 4^{\frac{3}{2}}, so you could rewrite 8^y as 8^y = 4^{\frac{3y}{2}}.

Can you see where to go on from there, if both sides are in the same form, you can simply equate the indices and turn it into a simple linear equation.

The second equation, you need to generally aim to turn the equation into two log terms. In this case, you'd want to turn the 4 into a log term. We know that \log_{2} 2^4 = 4, so you can rewrite 4 as that. Combine the two logs using the multiplication rule (as you probably know it), then you can simply 'cancel' out the logs and write it as a simple linear equation. I use cancel loosely, as that isn't what you're actually doing, but is a good approximation. Post your working and your two linear equations here, as well as any other questions you may have!
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TenOfThem
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(Original post by BBeyond)
I usually get most topics in maths, but I'm really struggling with logs for some reason

Take the attached question for example: Image

I don't even know where to begin on questions like this? I know all the rules but I just don't seem to be able to apply the right ones in the right situation?

Is there anyway to fix this besides just practicing because I don't really feel like I'm getting anywhere
I assume that you know (in terms of indices) what 8 and 4 have in common
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TeeEm
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(Original post by BBeyond)
I usually get most topics in maths, but I'm really struggling with logs for some reason

Take the attached question for example: Image

I don't even know where to begin on questions like this? I know all the rules but I just don't seem to be able to apply the right ones in the right situation?

Is there anyway to fix this besides just practicing because I don't really feel like I'm getting anywhere
there are 2 ways of doing this

the more sensible of the two is:

get rid of the logs in the 2nd equation
then solve 2 indicial equations
which eventually you can write as powers of 2
etc


EDIT: Gosh, very crowded here... I am out!
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BBeyond
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(Original post by Zacken)
x
Thank you for your help so far I have got to here now: http://gyazo.com/43662de50390114b8d9020041e41edb5 but am unsure how to proceed This may seem really dumb but can I just sub my y value from equation 1 into the log?

(Original post by TenOfThem)
I assume that you know (in terms of indices) what 8 and 4 have in common
Yeah I understand that, it's just the 2nd equation I'm mainly having problems with.

(Original post by TeeEm)
there are 2 ways of doing this

the more sensible of the two is:

get rid of the logs in the 2nd equation
then solve 2 indicial equations
which eventually you can write as powers of 2
etc

EDIT: Gosh, very crowded here... I am out!
Haha yeah got quite the response, thanks for your help anyway!
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Zacken
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(Original post by BBeyond)
Thank you for your help so far I have got to here now: http://gyazo.com/43662de50390114b8d9020041e41edb5 but am unsure how to proceed This may seem really dumb but can I just sub my y value from equation 1 into the log?



Yeah I understand that, it's just the 2nd equation I'm mainly having problems with.



Haha yeah got quite the response, thanks for your help anyway!
Ah, I see. That is correct.

We have that \log_{a} x = \log_{a} y \iff x=y, this is a rule that works when there are two log terms on opposite sides of the equation, of the same bases. Can you then see what the equation transforms into?
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BBeyond
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(Original post by Zacken)
Ah, I see. That is correct.

We have that \log_{a} x = \log_{a} y \iff x=y, this is a rule that works when there are two log terms on opposite sides of the equation, of the same bases. Can you then see what the equation transforms into?
Oh that was quite simple, thank you

What about this one? (if you don't mind ) http://gyazo.com/c02d02848068319ff4ae2711f31c62d7

This isn't my homework btw incase you think you're just doing my homework for me, I'm just trying to actually understand logs
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Zacken
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(Original post by BBeyond)
Oh that was quite simple, thank you

What about this one? (if you don't mind ) http://gyazo.com/c02d02848068319ff4ae2711f31c62d7

This isn't my homework btw incase you think you're just doing my homework for me, I'm just trying to actually understand logs
No, no, I don't think that, don't worry!

Yikes, I'm not having much luck with it. I'll keep trying, hopefully another member can do this better than I can.
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BBeyond
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(Original post by Zacken)
No, no, I don't think that, don't worry!

Yikes, I'm not having much luck with it. I'll keep trying, hopefully another member can do this better than I can.
No problem man, thanks anyway!

(Original post by TeeEm)
x
(Original post by TenOfThem)
x
If either of you two could help me with this question: http://gyazo.com/c02d02848068319ff4ae2711f31c62d7 I would much appreciate it
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TeeEm
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(Original post by BBeyond)
No problem man, thanks anyway!





If either of you two could help me with this question: http://gyazo.com/c02d02848068319ff4ae2711f31c62d7 I would much appreciate it
I sat, wrote the solution for you (because it is quite difficult to assist you on line and I am doing my own work as well) only to find messaging does not support attachments:mad:
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Zacken
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(Original post by BBeyond)
No problem man, thanks anyway!



If either of you two could help me with this question: http://gyazo.com/c02d02848068319ff4ae2711f31c62d7 I would much appreciate it

Okay, I've gotten it! It's a basic simultaneous equation.

We have the two equations:

a^x = b^y

a^x = (ab)^{xy}

How would you rewrite both of these equations in terms of logarithms?
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BBeyond
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(Original post by TeeEm)
I sat, wrote the solution for you (because it is quite difficult to assist you on line and I am doing my own work as well) only to find messaging does not support attachments:mad:
Oh damn Do you have a screen capture device or anything?

(Original post by Zacken)
Okay, I've gotten it! It's a basic simultaneous equation.

We have the two equations:

a^x = b^y

a^x = (ab)^{xy}

How would you rewrite both of these equations in terms of logarithms?
http://gyazo.com/1126dd63e164c98754d270450f1c0e6d

Am I right up until here?
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Zacken
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(Original post by BBeyond)
Oh damn Do you have a screen capture device or anything?



http://gyazo.com/1126dd63e164c98754d270450f1c0e6d

Am I right up until here?
You were fine upto x \log a = y \log b, leave this equation as it is for now.

Rewrite the second equation in terms of logs. (bring the power to the front of the log and divide by x throughout) What do you get then?
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TeeEm
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(Original post by Zacken)
You were fine upto x \log a = y \log b, leave this equation as it is for now.

Rewrite the second equation in terms of logs. (bring the power to the front of the log and divide by x throughout) What do you get then?
agreed
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BBeyond
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(Original post by Zacken)
You were fine upto x \log a = y \log b, leave this equation as it is for now.

Rewrite the second equation in terms of logs. (bring the power to the front of the log and divide by x throughout) What do you get then?
log(a)= y log (ab) ?

can i even divide by x? should it be x log(a) = xy log(ab)?
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Zacken
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(Original post by BBeyond)
log(a)= y log (ab) ?
Yes, great. Now how could you rewrite that y \log (ab) as two logs, using the addition/multiplication rule of logs?
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BBeyond
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(Original post by Zacken)
Yes, great. Now how could you rewrite that y \log (ab) as two logs, using the addition/multiplication rule of logs?
I can just divide by x like that? Doesn't seem very legit to me for some reason

but doing what you say I get log(a) = y log(a) + y log(b)

do i substitute into the first equation now?
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Zacken
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(Original post by BBeyond)
I can just divide by x like that? Doesn't seem very legit to me for some reason

but doing what you say I get log(a) = y log(a) + y log(b)

do i substitute into the first equation now?
If you have x \log a = xy \log b, then does't it seem rather natural that you could do \frac{x}{x} \log a= \frac{xy}{x} \log b?

Great, you again have that \log a = y \log a + y \log b, your first equation states that y \log b = x \log a, so you can rewrite the above equation as...?
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BBeyond
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(Original post by Zacken)
If you have x \log a = xy \log b, then does't it seem rather natural that you could do \frac{x}{x} \log a= \frac{xy}{x} \log b?

Great, you again have that \log a = y \log a + y \log b, your first equation states that y \log b = x \log a, so you can rewrite the above equation as...?
Yeah it does but then how come I couldn't do it for the first equation: http://gyazo.com/1126dd63e164c98754d270450f1c0e6d ?

Substituting in you get x log(a) = log(a)-y log(a)
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Zacken
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(Original post by BBeyond)
Yeah it does but then how come I couldn't do it for the first equation: http://gyazo.com/1126dd63e164c98754d270450f1c0e6d ?

Substituting in you get x log(a) = log(a)-y log(a)
You could, but it would just make your life really difficult and messy!

You have \log a = y \log a + y \log b, right?

Since x\log a = y\log b, the above equation becomes:

\log a = y \log a + x \log a \Rightarrow \log a = \log a^y + \log a^x, can you then combine this into an equation with two logs, then cancel the logs out?
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