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Logarithms

(b) Calculate the value of y for which 2 log3 y log3 ( y + 4 ) = 2 .
Here's my working out:

2 log3 y - log3 y - log3 4 = 2

log3 y - log3 4 = 2

log3 (y/4)=2

3^2=y/4

9=y/4

y=36

The correct answer is y=12. WHY isn't the above correct?? What have I done wrong?

THANKS

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Reply 1
Original post by krisshP
(b) Calculate the value of y for which 2 log3 y log3 ( y + 4 ) = 2 .
Here's my working out:

2 log3 y - log3 y - log3 4 = 2

log3 y - log3 4 = 2

log3 (y/4)=2

3^2=y/4

9=y/4

y=36

The correct answer is y=12. WHY isn't the above correct?? What have I done wrong?

THANKS


log3(y+4)log3y+log34log_{3} (y+4) \not= log_{3}y + log_{3}4

As loga(xy)=logax+logay log_{a} (xy) = log_{a}x + log_{a}y.

Posted from TSR Mobile
(edited 10 years ago)
Reply 2
Original post by krisshP
(b) Calculate the value of y for which 2 log3 y log3 ( y + 4 ) = 2 .
Here's my working out:

2 log3 y - log3 y - log3 4 = 2

log3 y - log3 4 = 2

log3 (y/4)=2

3^2=y/4

9=y/4

y=36

The correct answer is y=12. WHY isn't the above correct?? What have I done wrong?

THANKS


The line indicated
Reply 3
Original post by Joshmeid
log3(y+4)log3y+log34log_{3} (y+4) \not= log_{3}y + log_{3}4


Posted from TSR Mobile


Can you explain why please? I need to actually understand. Why can't you just expand the brackets?
Reply 4
Original post by krisshP
Can you explain why please? I need to actually understand. Why can't you just expand the brackets?


They are not equal

What do you want to understand

The log function will not work in that way
Reply 5
Original post by krisshP
Can you explain why please? I need to actually understand. Why can't you just expand the brackets?


As loga(xy)=logax+logay log_{a} (xy) = log_{a}x + log_{a}y

loga(x+y)logax+logaylog_{a}(x + y) \not= log_{a}x + log_{a}y


Posted from TSR Mobile
(edited 10 years ago)
Reply 6
Okay, thanks :smile:
Original post by Joshmeid
As loga(xy)=logax+logay log_{a} (xy) = log_{a}x + log_{a}y

loga(x+y)logax+logaylog_{a}(x + y) \not= log_{a}x + log_{a}y


Posted from TSR Mobile


Interestingly, in the case a=3, there is a value of m and a value of n such that

log3m+log3n=log3(m+n)\log_3 m + \log_3 n = \log_3 (m+n)

log3(m2)=(log3m)2\log_3 (m^2) = (\log_3 m)^2
(edited 10 years ago)
Reply 8
Original post by Indeterminate
...


Bad ... Indeterminate ... Bad
Original post by TenOfThem
Bad ... Indeterminate ... Bad


What? Both aren't generally true, yes, but what I've said is correct for a specific value of m and a specific value of n.
Original post by Indeterminate
What? Both aren't generally true, yes, but what I've said is correct for a specific value of m and a specific value of n.


KrisshP needs no confusion
Original post by TenOfThem
KrisshP needs no confusion


:tongue:

A bit too cheeky, maybe.
Reply 12
Original post by Indeterminate
Interestingly, in the case a=3, there is a value of m and a value of n such that

log3m+log3n=log3(m+n)\log_3 m + \log_3 n = \log_3 (m+n)

log3(m2)=(log3m)2\log_3 (m^2) = (\log_3 m)^2


If it's any consolation, I thought it was pretty cool too. Shhh, don't tell ToT
Original post by Occams Chainsaw
Shhh, don't tell ToT


I can read LoL
Original post by krisshP
Can you explain why please? I need to actually understand. Why can't you just expand the brackets?

It's because a logarithm is a function - in the same way (2+3)^2 is not the same as (2^2) + (3^2). :smile:
Reply 15
Original post by TenOfThem
I can read LoL


I figured you would only come back to the thread if you were quoted! :tongue:
Original post by Occams Chainsaw
I figured you would only come back to the thread if you were quoted! :tongue:


I can smell out my name even if it is abbreviated :colone:
Okay, so here is my working:

2log3 y - log3(y+4) = 2
log3 y^2 - log3(y+4) = 2
log3( y^2 / y+4 ) = 2
Take 3 to the power of both sides to give
3^ (log3 y^2 / y+4 ) = 3^2
y^2 / y+4 = 9
multiply through by y + 4
y^2 = 9 (y+4)
y^2 = 9y = 36
y^2 - 9y -36 = 0
(y - 12) (y + 3) = 0
so y = 12, -3
can't do a negative number with logs, so y = 12 is the answer

Hope this helped :smile:
(edited 10 years ago)
Reply 18
Original post by TenOfThem
I can smell out my name even if it is abbreviated :colone:


I've got a good log question that my maths lectures couldn't/wouldn't answer for me so I'm wondering if you can/will please!

Why do negative base values sometimes work but sometimes not? I presume it's got something to do with complex numbers but I can't make the connection tbh.

Sooooo many things annoy me about A Level maths like this! Like why the CoM of a semi-circle is 4(r)/3(pi) and why exactly the way we integrating (namely dividing my the new power) actually works. Or how sine is actually derived (not that circle on an axis thing, that's cheating!).

Mechanical use of the principles is stupid.. We need to know why, I think! Maybe I should be able to work it out for myself but I am far too lazy and probably not clever enough to do it!
(edited 10 years ago)
Original post by Occams Chainsaw
I've got a good log question that my maths lectures couldn't/wouldn't answer for me so I'm wondering if you can/will please!

Why do negative base values sometimes work but sometimes not? I presume it's got something to do with complex numbers but I can't make the connection tbh.

Sooooo many things annoy me about A Level maths like this! Like why the CoM of a semi-circle is 4(r)/3(pi) and why exactly the way we integrating (namely dividing my the new power) actually works. Or how sine is actually derived (not that circle on an axis thing, that's cheating!).

Mechanical use of the principles is stupid.. We need to know why, I think! Maybe I should be able to work it out for myself but I am far too lazy and probably not clever enough to do it!


Well, to answer your main Q

log39=2\log_{-3} 9 = 2

works but it's not of any use tbh
(edited 10 years ago)

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