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Removing logs from both sides

Page 7 question 8 last part

How do you remove the logbase10 from both sides?
Reply 1
Avatar for BananaPie
BananaPie
You're missing an attachment. =S
Original post by BananaPie
You're missing an attachment. =S


http://www.ocr.org.uk/Images/59336-mark-scheme-january.pdf
Reply 3
Avatar for Acruzen
Acruzen
10
you just do :/ they are the same on both sides so you can just cancel?
Reply 4
Avatar for Konflict
Konflict
7
Divide both sides by log10. The x^2-10/9 could be in brackets, to make it easier to see.
Reply 5
Avatar for BananaPie
BananaPie
Logbase10 ((x^2-10)/x) = logbase10 (9)

Since they're both logbase10 with no other constants or variables added to the equated equation, (x^2-10)/x has to equal 9 in order for this equated equation to work. Hence when you solve it you do not need to bother with Logbase10.

If you're given a situation where logbase10 (2a) = logbase10 (5), then 2a = 5. However if logbase10 (2a) = logbase10 (5) + 3, then you cannot say that 2a = 5. You'll have to convert 3 into a logarithm with base 10.
(edited 11 years ago)
Original post by Acruzen
you just do :/ they are the same on both sides so you can just cancel?



Original post by Konflict
Divide both sides by log10. The x^2-10/9 could be in brackets, to make it easier to see.


But log 10 / log 10 = syntax error and not 1 ?
Original post by Konflict
Divide both sides by log10. The x^2-10/9 could be in brackets, to make it easier to see.


Really no.

The inverse operation of taking logs to base 10 is to exponentiate to base 10.
Original post by BananaPie
Logbase10 ((x^2-10)/x) = logbase10 (9)

Since they're both logbase10 with no other constants or variables added to the equated equation, (x^2-10)/x has to equal 9 in order for this equated equation to work. Hence when you solve it you do not need to bother with Logbase10.


Yes thanks I remember my teacher telling me this :smile:
Original post by IShouldBeRevising_
But log 10 / log 10 = syntax error and not 1 ?


log 10 / log 10 = 1

but that is irrelevant here as that is not the required approach.
Reply 10
Avatar for Konflict
Konflict
7
Original post by IShouldBeRevising_
But log 10 / log 10 = syntax error and not 1 ?


Sorry, look at BananaPie's post! :redface:
It uses log property: logb(b)^x=x (I think, Mr M will correct me! :tongue:)
So, log10(10^9)=x^2-10/x
log10(10^9)=9
9=x^2-10/x

I don't know why I said what I said. :redface: Just ignore me.
(edited 11 years ago)
Original post by Mr M
log 10 / log 10 = 1

but that is irrelevant here as that is not the required approach.


Yes thank you, I understand now.

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