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1. Log question - express in terms of logax , logay, logaz

Ive attached the ques h.) & my working but i dont understand where the 5/2 comes from.

Answer is 5/2 + 1/2logax - 3logay + 5/2logaz

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Attached Images

2. di
(Original post by kiiten)
Log question - express in terms of logax , logay, logaz

Ive attached the ques h.) & my working but i dont understand where the 5/2 comes from.

Answer is 5/2 + 1/2logax - 3logay + 5/2logaz

Posted from TSR Mobile

Now

and

Can you take it from here? What's ?
3. (Original post by Zacken)
di

Now

and

Can you take it from here? What's ?
How did you get to -logay^3 on the second line of working ?
4. (Original post by kiiten)
How did you get to -logay^3 ?
except in this case, we have and
5. (Original post by Zacken)
except in this case, we have and
Umm i think ive done it wrong but i got
2 + 1/2loga + 1/2logx + 5/2logz - 3logay
6. (Original post by kiiten)
Umm i think ive done it wrong but i got
2 + 1/2loga + 1/2logx + 5/2logz - 3logay
Yeah, that's correct, just remember that so you can simplify your first two terms as
7. (Original post by Zacken)
Yeah, that's correct, just remember that so you can simplify your first two terms as
So 5/2 + 1/2logax + 5/2logaz - 3logay
8. (Original post by kiiten)
So 5/2 + 1/2logax + 5/2logaz - 3logay
Seems fine, yups.
9. Thank you

If you have log (1+5^-x) would it simplify to -xlog6 ?
10. (Original post by kiiten)
Thank you

If you have log (1+5^-x) would it simplify to -xlog6 ?
No, it wouldn't; you can't simplify that anymore. Remember that . :-)
11. (Original post by Zacken)
No, it wouldn't; you can't simplify that anymore. Remember that . :-)
So how would you solve
Log (1×5^-x) = log3
12. (Original post by kiiten)
So how would you solve
Log (1×5^-x) = log3
"cancel" the logs: - now you know how to solve this using logarithms.
13. (Original post by Zacken)
"cancel" the logs: - now you know how to solve this using logarithms.
Oh i see - could you do that for any log e.g. log5^-x = log3 instead of simplifying
14. (Original post by kiiten)
Oh i see - could you do that for any log e.g. log5^-x = log3 instead of simplifying
If you have - then you know that .

If you have IT IS NOT TRUE THAT BUT IT IS TRUE that

Basically when you have log equations you should try and get it in the form then you can say something = something else. (the bases of the logs have to be the same, if they're not, you can always use change of base rule) .
15. (Original post by Zacken)
If you have - then you know that .

If you have IT IS NOT TRUE THAT BUT IT IS TRUE that

Basically when you have log equations you should try and get it in the form then you can say something = something else. (the bases of the logs have to be the same, if they're not, you can always use change of base rule) .
Thanks that helps to explain things
16. (Original post by kiiten)
Thanks that helps to explain things
No problem.
17. What about this question - im not sure where you would start. Ive attached some workingAttachment 513103513107

Attached Images

18. Use Log rules.
19. (Original post by Mystery.)

Use Log rules.
Have i done it right so far. So, 5/ (log3x) = 6 / (log3)
20. (Original post by kiiten)
What about this question - im not sure where you would start. Ive attached some workingAttachment 513103513107

You have

Let and then you have

So you can ignore logs and solve these equations for and . Then substitute back in for and .

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