The Student Room Group

Maths logarithms help!!

Show that... log(base a)x^10 - 2log(base a)(x^3/4) = 4log(base a)(2x)
I can only get close to the answer but not the actual I don't think I fully understand the rules. Help would be appreciated my exams on Friday :frown:


Posted from TSR Mobile
Reply 1
Original post by Bethj1234
Show that... log(base a)x^10 - 2log(base a)(x^3/4) = 4log(base a)(2x)
I can only get close to the answer but not the actual I don't think I fully understand the rules. Help would be appreciated my exams on Friday :frown:


Posted from TSR Mobile


Do you know the following rules:

alogcb=logcbaa\log_c b = \log_c b^a

logcalogcb=logcab\log_c a - \log_c b = \log_c \frac{a}{b}
Reply 2
Yes I do! But I can't seem to apply them correctly to this question


Posted from TSR Mobile
Reply 3
Original post by Bethj1234
Yes I do! But I can't seem to apply them correctly to this question


Posted from TSR Mobile


First, try applying

alogcb=logcbaa\log_c b = \log_c b^a

to all the terms in the equation.
Reply 4
Original post by Bethj1234
Show that... log(base a)x^10 - 2log(base a)(x^3/4) = 4log(base a)(2x)
I can only get close to the answer but not the actual I don't think I fully understand the rules. Help would be appreciated my exams on Friday :frown:


Posted from TSR Mobile


Wait. I think you have not typed the question correctly. Try substituting a value for x and LHS is not equal to RHS.
Reply 5
Original post by Vorte
Wait. I think you have not typed the question correctly. Try substituting a value for x and LHS is not equal to RHS.


You're supposed to find a value for x, obviously substituting a random x value isn't going to work...
Reply 6
I have done but I end up with log(base a)x^10 - log(base a)(x^6/16). It's he fraction I'm struggling with. I tried to make it not a fraction like you'd do with 1/x^2 becoming x^-2 but I'm not sure how to do it x on top.


Posted from TSR Mobile
Reply 7
Does this work? I've probably done it wrong... ;L

Where did you find this question? Seems too complex to come up on Friday.
Reply 8
I think you may have read part of the question wrong, also it only says show that! Solving for x would be too nasty to come up. Here's the question


Posted from TSR Mobile
Reply 9
Oh and the question is from june 2007 paper!


Posted from TSR Mobile
Reply 10
Here you are:
Original post by spiruel
Here you are:


Why has the sign changed in the second step from "-" to a "+"
Reply 12
Original post by Hi, How are you ?
Why has the sign changed in the second step from "-" to a "+"


Log rule - the minus 2 has been put as a power for (x^3/4). You have you interpret minuses as plussing a minus... If that makes sense. Basically, revise log rules so you're sure. :wink:

Posted from TSR Mobile
Reply 13
Original post by Bethj1234
I think you may have read part of the question wrong, also it only says show that! Solving for x would be too nasty to come up.



You didn't put brackets around the X^3/4 in your first post. That's why I misread it! You should have wrote it as (x^3)/4. :smile:

Posted from TSR Mobile
Reply 14
Ah it's so simple thank you! I didn't realise if you had a coefficient infront of the x that you had to take the 4th root of 16 to make it 4loga(2x) thought it was just left as 16


Posted from TSR Mobile

Quick Reply

Latest