Competition - post a photo you've taken of nature >>

Logs help

I usually get most topics in maths, but I'm really struggling with logs for some reason :frown:

Take the attached question for example:

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 :frown:

(edited 9 years ago)

edf4932d8926504774c2d47b24bdb67e.png151.6KB

Scroll to see replies

Reply 1

Zacken

Original post by BBeyond

I usually get most topics in maths, but I'm really struggling with logs for some reason :frown:

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 :frown:

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

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! :smile:

Reply 2

TenOfThem

Original post by BBeyond

I usually get most topics in maths, but I'm really struggling with logs for some reason :frown:

Take the attached question for example:

I assume that you know (in terms of indices) what 8 and 4 have in common

Reply 3

TeeEm

Original post by BBeyond

I usually get most topics in maths, but I'm really struggling with logs for some reason :frown:

Take the attached question for example:

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!

(edited 9 years ago)

Reply 4

BBeyond

Original post by Zacken

Thank you for your help so far :smile:

I have got to here now: http://gyazo.com/43662de50390114b8d9020041e41edb5 but am unsure how to proceed :frown:

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

Haha yeah got quite the response, thanks for your help anyway!

Reply 5

Zacken

Original post by BBeyond

Thank you for your help so far :smile:

I have got to here now: http://gyazo.com/43662de50390114b8d9020041e41edb5 but am unsure how to proceed :frown:

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? :smile:

Reply 6

BBeyond

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? :smile:

Oh that was quite simple, thank you :biggrin:

What about this one? (if you don't mind :tongue:

) 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 :smile:

Reply 7

Zacken

Original post by BBeyond

Oh that was quite simple, thank you :biggrin:

What about this one? (if you don't mind :tongue:

) 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 :smile:

No, no, I don't think that, don't worry! :tongue:

Yikes, I'm not having much luck with it. I'll keep trying, hopefully another member can do this better than I can. :redface:

Reply 8

BBeyond

Original post by Zacken

No, no, I don't think that, don't worry! :tongue:

Yikes, I'm not having much luck with it. I'll keep trying, hopefully another member can do this better than I can. :redface:

No problem man, thanks anyway! :smile:

Original post by TeeEm

Original post by TenOfThem

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

Reply 9

TeeEm

Original post by BBeyond

No problem man, thanks anyway! :smile:

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

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:

Reply 10

Zacken

Original post by BBeyond

No problem man, thanks anyway! :smile:

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

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

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? :smile:

Reply 11

BBeyond

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. :smile:

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? :smile:

http://gyazo.com/1126dd63e164c98754d270450f1c0e6d

Am I right up until here?

Reply 12

Zacken

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? :smile:

Reply 13

TeeEm

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? :smile:

agreed

Reply 14

BBeyond

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? :smile:

log(a)= y log (ab) ?

can i even divide by x? should it be x log(a) = xy log(ab)?

(edited 9 years ago)

Reply 15

Zacken

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? :smile:

Reply 16

BBeyond

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? :smile:

I can just divide by x like that? Doesn't seem very legit to me for some reason :frown:

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

do i substitute into the first equation now?

Reply 17

Zacken

Original post by BBeyond

I can just divide by x like that? Doesn't seem very legit to me for some reason :frown:

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...? :smile:

Reply 18

BBeyond

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...? :smile:

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)

Reply 19

Zacken

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)