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LOGARITHMS homework help

can somebody please help me to understand questions f-j? I'm frustrated on log, and I'm unsure on how to punch the numbers into the calculator. please help :frown:(
(edited 9 years ago)
Reply 1
Original post by kandykissesxox
can somebody please help me to understand questions f-j? I'm frustrated on log, and I'm unsure on how to punch the numbers into the calculator. please help :frown:(


Did you manage to do part (e)? Part (f) uses exactly the same principles of taking logs and solving a linear equation for x.
Original post by kandykissesxox
..

Have you attempted the question by any chance? If so can you show your workings :smile:
Reply 3
pretty simple take log both sides
Reply 4
g) use power law

then you should each until here-->

(x-1) = log 3
(x+1) log 8

solve and simply from there
to get x = 3.24..
(edited 9 years ago)
Original post by Sena5
F)

log 7^(2x-1) = log 23

log (2x-1) = 1.36...
log 7

log (2x-1) = 1.36.. * (log 7)
log 2x-1 = 1.15...
log 2x = 2.15...

10^(2.15..) = x
2

:. x = 70.75...


This is complete nonsense.
72x1=23    (2x1)log7=log23    2x1=log23log77^{2x-1}=23\implies (2x-1)log7=log23\implies 2x-1=\frac{log23}{log7}
Original post by Sena5
g) use power law

then you should each until here-->

(x-1) = log 3
(x+1) log 8

solve and simply from there
to get x = 3.24..


This is totally wrong again.
(edited 9 years ago)
Reply 7
Original post by Sena5
F)

log 7^(2x-1) = log 23

log (2x-1) = 1.36...
log 7

log (2x-1) = 1.36.. * (log 7)
log 2x-1 = 1.15...
log 2x = 2.15...

10^(2.15..) = x
2

:. x = 70.75...


Are you trolling the Maths forum now, because this is the second pile of rubbish you've posted in a thread in the last 24 hours!

(And in any case, it is against forum rules to post full solutions!)
Original post by Sena5
F)

log 7^(2x-1) = log 23

log (2x-1) = 1.36...
log 7

log (2x-1) = 1.36.. * (log 7)
log 2x-1 = 1.15...
log 2x = 2.15...

10^(2.15..) = x
2

:. x = 70.75...


lmao u wot?

log7^(2x-1) = log23

(2x-1)log7 = log23

(2x-1) = log23/log7

etc.

Type the logarithm power rule in to Google.
(edited 9 years ago)

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