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2. (Original post by kandykissesxox)
Did you manage to do part (e)? Part (f) uses exactly the same principles of taking logs and solving a linear equation for x.
3. (Original post by kandykissesxox)
..
Have you attempted the question by any chance? If so can you show your workings
4. pretty simple take log both sides
5. 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..
6. (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.
7. (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.
8. (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!)
9. (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.

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