… continued from the previous post
“a=2! b=1! c=3!“, Miss Loi cried out at the last moment, just as the mighty wall of water towered over her, and then everything went blank …
She awoke to a scene of utter desolation. Crucially, a glance downwards at her torn and tattered bikini convinced her that 此地不可久留, as she’d literally be ‘easy meat’ for the scores of horny cheekopeks* roaming the area.
For that, she needed to quickly find her way back to The Temple. But she now stood in an unrecognizable landscape where everything was swept away by the waves, save for a wooden log and a thick power cable on the ground.
Upon closer inspection, writings were found on each of them but with crucial parts of the details missing.
Inscribed on the wooden log was this:
Given: logb(xy3) = m and logb(x3y2) = p
Distance to The Temple (km) from here = logb√(xy).
To find this distance, first express it in terms of m and p …
*workings destroyed*
… and finally substitute in m=1 and p=2 .
Miss Loi turned over the wooden log and found this:
The Laws of Logarithms
- logbxy = logbx + logby
- logb(x/y) = logbx – logby
- logbxn = nlogbx
And printed on the thick power cable was this:
Given: 22x+2 x 5x-1 = 8x x 52x.
Length of cable (km) from here to The Temple = 10x …
*workings destroyed*
Some graffiti was also found on the cable:
The Rules of Powers/Indices
- am x an = am+n
- am ÷ an = am-n
- (am)n = amn
- a-n = 1/an
- (a x b)n = an x bn
At that very moment, packs of creatures looking like this appeared suddenly on the horizon, and were closing in fast on Miss Loi!
In which direction should she flee? In the direction of the log or the power cable?!
* Local term for dirty old man
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Miss Loi is a full-time private tutor in Singapore specializing in O-Level Maths tuition. Her life's calling is to eradicate the terrifying LMBFH Syndrome off the face of this planet. For over 21 years she has been a savior to countless students ... 
12 Comments
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Wah Laueh! Aiyah! Miss! Me where got got so bad wan! Don worry I wouldn't simply 'eat' u wan ( pokeing is another matter )! He! He! Oh! There is a S'pore movie by this name from jack Neo! A very nice movie! Have u watch it? BTW what is a 'cheekopeks' ??? Have a nice weekend!
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HAM, you really don't know what cheekopek is?! You CHEE-KO-PEK!
*Quickly runs and hide behind the log*
曜
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erm i dun get what the quetion means>.< both parts are different or connection? for part 2 i got
2^(2x + 2) x 5^(x-1) = 2^(3x) x 5^(2x)
by comparison 2x + 2 = 3x
x = 2
or 2x = x-1
x= 1
doesnt make sense-.-
曜
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Ok kiroii to cut a long story short:
FOR PART 1,
Distance to The Temple = logb√(xy)
Given that : logb(xy3) = m and logb(x3y2) = p,
Express logb√(xy) in terms of m and p.
FOR PART 2,
Distance to The Temple = 10x
Given that 22x+2 x 5x-1 = 8x x 52x,
Find the value of 10x.
Note: The final answer for 10x is numerical, there shouldn't be any more x present.
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erm fer the first part it the value of m and p given? if not i am truely clueless>.
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er is my comment partially deleted? coz thats not what i've posted but if its the filter do bring it up if not let me know i'll repost as i got the solution for part 2
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Part 2
22x+2 x 5x-1=23x x 52x
5(x-1-2x)=2(3x-2x-2)
5(-x-1)=2(x-2)
5(-1) x 22=2x x 5x
Therefore, 10x=4/5
That's right. From one of the rules of indices you know that 10x can be expressed in terms of 2x x 5x, so you'll need to do is to get the expression in terms of 2s and 5s. Hence the first step is to make 8 into 23.
From this point onwards, there're many ways where you can manipulate the expression using your indices wizardry.
Tears of anxiety plummeted the prodigy's cheek, as his efforts to solve Part 1 seems to be futile.
曜
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Since today's a special day (i.e. the grand O-Level eve), and to stem the flow of someone's tears, Miss Loi shall be nice and put up the solution to Part 1:
When you see a weird expression like logb√(xy), you can almost be sure that you'll need to expand it using your Laws of Logarithm. So,
logb√(xy) = 1/2(logbx + logby)
So now you know you'll simply have to find logbx and logby.
Now looking at the two given expressions containing both x and y, you can easily express logbx and logby in terms of m and p by forming two simultaneous equations using some Laws of Logarithm wizardry:
log(xy3) = m
logbx + logby3 = m
logbx + 3logby = m
logbx = m - 3logby ---- (1)
logbx3y2 = p
logbx3 + logby2 = p
3logbx + 2logby = p ---- (2)
Sub (1) into (2) to get logby
3(m - 3logby) + 2logby = p
3m - 9logby + 2logby = p
-7logby = p - 3m
logby = (3m-p)/7 ---- (3)
Sub (3) back into (1) to get logbx
logbx = m - [3(3m-p)/7]
logbx = (7m-9m+3p)/7
logbx = (3p-2m)/7
∴ logb√(xy) = 1/2(logbx + logby)
= 1/2[(3p-2m)/7 + (3m-p)/7 ]
= 1/2[(m+2p)/7]
= (m+2p)/14
So given that m=1 and p=2, which is the shortest way back to The Temple?
曜
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w00t 6hrs till showtime gl to all those who are taking a maths >.>
曜
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Good luck kiroii! Read your questions carefully!
曜
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Given m=1 and p=2, logb√(xy) = [1+2(2)]/14 = 5/14 < 4/5
.'. The shortest way back to the temple is from the wooden log.
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hahaha 2007 blog still helps