[jgupta]

Hi - I'm revising for the C2 exam and logarithms are pretty much the last thing I have to get sorted....

I get the basic stuff ( subtract, add, divide, multiply logs etc) but don't get this question:

(I have removed a from log a for ease of reading)

"Show that logX^10 - 2log(x^3 / 4) = 4log(2x)"

My process so far has been:

10logX - (2logX^3 - 2log4) = 4log(2x)
10logX - (6logX - 2log4) = 4log(2x)

but I can't go from here...and don't even know if I'm doing that right...If someone could help me out I'd really appreciate it.

Thanks
James

[Adampolar]

log(ab)=loga + logb use that on the right hand side.

[Number6]

Try not splitting up the second term of the LHS and then square all of it.

[brockham]

is it X to the power of 3/4... or X to the power of 3.. all over 4?

[rye225]

stuck on this same question lol, was just about to post.

it is x cubed / 4

[rye225]

still having trouble, anyone got a worked answer?

[Number6]

I think this is right...

logx^10 - 2log((x^3)/4)
logx^10 - log((x^6)/16)
log(x^10 /((x^6)/16)))
log ((16x^10)/x^6)
log16x^4
4log2x

[jgupta]

Thanks for all the replies

so for 4log(2x)...this can be written as:

4log2 + 4logx ?

[moody28028]

Originally Posted by Number6

I think this is right...

logx^10 - 2log((x^3)/4)
logx^10 - log((x^6)/16)
log(x^10 /((x^6)/16)))
log ((16x^10)/x^6)
log16x^4
4log2x

We went through this question in maths on friday, and there were various methods, but this is the one I wrote down.

[rye225]

thanks!!

[Adampolar]

Originally Posted by jgupta

Thanks for all the replies

so for 4log(2x)...this can be written as:

4log2 + 4logx ?

yep, im glad someone took my hint.

[jgupta]

Thanks guys for all your help - i've solved it now

Seems like this is a pretty infamous question lol

[jgupta]

Originally Posted by Angus-Higgins

Solution

.

Angus Higgins

Hi - I like your solution because it follows how I did it up to where I got stuck...and I understand it all now except for the last part:

2log4X^2
4log2X

how do you go from the first line to the second? I would have thought that the ^2 multiplies the first "2" to 4log...but then it is still 4X ant not 2X?

[sohanshah]

Originally Posted by jgupta

Hi - I like your solution because it follows how I did it up to where I got stuck...and I understand it all now except for the last part:

2log4X^2
4log2X

how do you go from the first line to the second? I would have thought that the ^2 multiplies the first "2" to 4log...but then it is still 4X ant not 2X?

Remember your log rules?