# logsWatch

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#1
Can someone help explain them to me, i dont understant them at all

E.g 3^x = 5

Id be grateful for a guided answer, thank you!
0
14 years ago
#2
(Original post by LondonBoy)
Can someone help explain them to me, i dont understant them at all

E.g 3^x = 5

Id be grateful for a guided answer, thank you!
take logs

xlog3 = log 5 (the x comes down in front of the log)
x = log 5 / log 3 (rearrange)

x = 1.46...
0
14 years ago
#3
(Original post by LondonBoy)
Can someone help explain them to me, i dont understant them at all

E.g 3^x = 5

Id be grateful for a guided answer, thank you!
Take logs of both sides.
log3^x = log5
loga^b = bloga
xlog3 = log5
x = log5/log3
0
14 years ago
#4
(Original post by manps)
take logs

xlog3 = log 5 (the x comes down in front of the log)
x = log 5 / log 3 (rearrange)

x = 1.46...
hey manps.
On www.mrmaths.co.uk I think you should move the m1-m3 jan 05 papers to the new specification page as it would be easier to view them if that's okay?
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#5
0
14 years ago
#6
Might just be worth noting that the answer

log5/log3

that people are getting can also be written

log_3 5

that is - log to base 3 of 5.

The inverse of "x -> 3^x" is "y -> log_3 y".

I guess the log, that people are writing, means to base 10 - but there's nothing special about base 10 and we could equally have "taken logs of both sides to base 3".
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14 years ago
#7
(Original post by Widowmaker)
hey manps.
On www.mrmaths.co.uk I think you should move the m1-m3 jan 05 papers to the new specification page as it would be easier to view them if that's okay?
Hi
Ive got someone designing the site at the moment. I'm putting all my effort into that at the moment. Haven't got much time for updates but keep your eyes peeled for when the new site appears online. It's quite a nice design and is much easier to nagvigate than the current site.

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