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2E: Laws of logarithms, (3) (c) (i)

1455368728178-271472226.jpg

Help with (c) (i) please :puppyeyes:

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Original post by IFoundWonderland
1455368728178-271472226.jpg

Help with (c) (i) please :puppyeyes:

Darling, dear, just change the base of the logarithm. Not that hard..

Ultimately, you get a quadratic using y-substitution which you can solve quite simply. What course are you doing? This is a joke.
(edited 8 years ago)
Original post by Bath_Student
Darling, dear, just change the base of the logarithm. Not that hard..


I haven't done this in about a year and am re self teaching for a resit from a shitty textbook. Idek what that means :bawling:
Original post by IFoundWonderland
1455368728178-271472226.jpg

Help with (c) (i) please :puppyeyes:


I wasn't going to help at first, but then I saw the puppyeyes ...

If log⁡4x=y\log_4 x = y then

4y=x⇒(22)y=x⇒22y=x⇒log⁡2x=2y=2log⁡4x4^y = x \Rightarrow (2^2)^y =x \Rightarrow 2^{2y}=x \Rightarrow \log_2 x = 2y = 2 \log_4 x
Original post by IFoundWonderland
1455368728178-271472226.jpg

Help with (c) (i) please :puppyeyes:


try these on your calculator and see if it gives you some clues

log base 3 of 3

log base 1.4 of 1.4
Original post by IFoundWonderland
I haven't done this in about a year and am re self teaching for a resit from a shitty textbook. Idek what that means :bawling:


Alright! :wink:

So, change log(4) x into base 2, gives you log(2) x / log(2) 4. It's a rule, learn it.
log(2) 4 is 2. So, you get that log(4) x = log(2)x / 2. Multiply through by 2 to remove the 2 in the denominator. Now, then, the other log(2) x is doubled (as you multiplied through by 2), and a law of logs states that 2*log(2) x = log(2) x^2.

Now you have a quadratic, let y = log(2) x. Got it from there?!
Original post by Bath_Student
Darling, dear, just change the base of the logarithm. Not that hard..

Ultimately, you get a quadratic using y-substitution which you can solve quite simply. What course are you doing? This is a joke.


that is a very unhelpful comment.
Original post by atsruser
I wasn't going to help at first, but then I saw the puppyeyes ...

If log⁡4x=y\log_4 x = y then

4y=x⇒(22)y=x⇒22y=x⇒log⁡2x=2y=2log⁡4x4^y = x \Rightarrow (2^2)^y =x \Rightarrow 2^{2y}=x \Rightarrow \log_2 x = 2y = 2 \log_4 x

Are you drunk?
Original post by Bath_Student
Darling, dear, just change the base of the logarithm. Not that hard..

Ultimately, you get a quadratic using y-substitution which you can solve quite simply. What course are you doing? This is a joke.

I do the IB.

Original post by atsruser
I wasn't going to help at first, but then I saw the puppyeyes ...

If log⁡4x=y\log_4 x = y then

4y=x⇒(22)y=x⇒22y=x⇒log⁡2x=2y=2log⁡4x4^y = x \Rightarrow (2^2)^y =x \Rightarrow 2^{2y}=x \Rightarrow \log_2 x = 2y = 2 \log_4 x


Original post by maggiehodgson
try these on your calculator and see if it gives you some clues

log base 3 of 3

log base 1.4 of 1.4

1455369639286-440974274.jpg
The correct answer is 1/4. Also it is a non calculator question *sigh*
Original post by IFoundWonderland






The correct answer is 1/4. Also it is a non calculator question *sigh*

AHHH!!!!!!!!!!!

I reiterate:

Alright!

So, change log(4) x into base 2, gives you log(2) x / log(2) 4. It's a rule, learn it.
log(2) 4 is 2. So, you get that log(4) x = log(2)x / 2. Multiply through by 2 to remove the 2 in the denominator. Now, then, the other log(2) x is doubled (as you multiplied through by 2), and a law of logs states that 2*log(2) x = log(2) x^2.

Now you have a quadratic, let y = log(2) x. Got it from there?!
Original post by Bath_Student
Darling, dear, just change the base of the logarithm. Not that hard..

Ultimately, you get a quadratic using y-substitution which you can solve quite simply. What course are you doing? This is a joke.


no need to be so patronising an unpleasant :erm:

good luck wonderland :lovehug: xxx
Original post by IFoundWonderland
I do the IB.


In the UK?
Original post by IFoundWonderland
I do the IB.




1455369639286-440974274.jpg
The correct answer is 1/4. Also it is a non calculator question *sigh*


Well, I was just trying to get you to experiment to see that log to base a of a is 1.

Form that you should be able to do something with line 1.
Original post by Bath_Student
When women ransack the mathematics thread! :smile:


how evil of them :afraid:
tbf i don't know anything about maths so i'll leave lol just came cos wonderland is here :h:
these ppl could help you tho wonderland lovely :smile: :lovehug:
@jamestg
@Student403
@TeeEm
@SeanFM
Original post by CoolCavy
how evil of them :afraid:
tbf i don't know anything about maths so i'll leave lol just came cos wonderland is here :h:
these ppl could help you tho wonderland lovely :smile: :lovehug:


Damnit I answered her question. If her IQ is larger then 3, then she'll be fine.
Original post by Bath_Student
AHHH!!!!!!!!!!!

I reiterate:

Alright!

So, change log(4) x into base 2, gives you log(2) x / log(2) 4. It's a rule, learn it.
log(2) 4 is 2. So, you get that log(4) x = log(2)x / 2. Multiply through by 2 to remove the 2 in the denominator. Now, then, the other log(2) x is doubled (as you multiplied through by 2), and a law of logs states that 2*log(2) x = log(2) x^2.

Now you have a quadratic, let y = log(2) x. Got it from there?!


Um

So when the x is squared its the equivalent of the whole log being squared?

I didn't realise that :/

Christ I'm so confused rn and my notes look so messy:frown:
Original post by the bear
that is a very unhelpful comment.


I already spent my bear rep™ today it seems… PRSOM.
Original post by IFoundWonderland
Um

So when the x is squared its the equivalent of the whole log being squared?

I didn't realise that :/

Christ I'm so confused rn and my notes look so messy:frown:


n*log a = log a^n. You must have seen that before?

Therefore, 2 log(2) x = log(2) x^2 .
Original post by CoolCavy
how evil of them :afraid:
tbf i don't know anything about maths so i'll leave lol just came cos wonderland is here :h:
these ppl could help you tho wonderland lovely :smile: :lovehug:
@jamestg
@Student403
@TeeEm
@SeanFM


:wavey:

:mad:
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TeeEm
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Original post by CoolCavy
how evil of them :afraid:
tbf i don't know anything about maths so i'll leave lol just came cos wonderland is here :h:
these ppl could help you tho wonderland lovely :smile: :lovehug:
@jamestg
@Student403
@TeeEm
@SeanFM


teaching at present

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