Comments on: Just Follow Law Or Just Follow Rule?

By: ang

ang — Thu, 22 Dec 2011 21:13:28 +0000

hahaha 2007 blog still helps

By: Li-sa

Li-sa — Sun, 14 Sep 2008 04:25:23 +0000

Given m=1 and p=2, log_b√(xy) = [1+2(2)]/14 = 5/14 < 4/5 .'. The shortest way back to the temple is from the wooden log.

By: Miss Loi

Miss Loi — Mon, 22 Oct 2007 07:21:59 +0000

Good luck kiroii! Read your questions carefully!

By: kiroii

kiroii — Mon, 22 Oct 2007 00:10:25 +0000

w00t 6hrs till showtime gl to all those who are taking a maths >.>

By: Miss Loi

Miss Loi — Sun, 21 Oct 2007 14:35:44 +0000

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 log_b√(xy), you can almost be sure that you'll need to expand it using your Laws of Logarithm. So, log_b√(xy) = 1/2(log_bx + log_by) So now you know you'll simply have to find log_bx and log_by. Now looking at the two given expressions containing both x and y, you can easily express log_bx and log_by in terms of m and p by forming two simultaneous equations using some Laws of Logarithm wizardry: log(xy³) = m log_bx + log_by³ = m log_bx + 3log_by = m log_bx = m - 3log_by ---- (1) log_bx³y² = p log_bx³ + log_by² = p 3log_bx + 2log_by = p ---- (2) Sub (1) into (2) to get log_by 3(m - 3log_by) + 2log_by = p 3m - 9log_by + 2log_by = p -7log_by = p - 3m log_by = (3m-p)/7 ---- (3) Sub (3) back into (1) to get log_bx log_bx = m - [3(3m-p)/7] log_bx = (7m-9m+3p)/7 log_bx = (3p-2m)/7 ∴ log_b√(xy) = 1/2(log_bx + log_by) = 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? :)

By: Prodigy

Prodigy — Sun, 21 Oct 2007 11:11:34 +0000

Part 2 2^2x+2 x 5^x-1=2^3x x 5^2x 5^(x-1-2x)=2^(3x-2x-2) 5^(-x-1)=2^(x-2) 5^(-1) x 2²=2^x x 5^x Therefore, 10^x=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.

By: kiroii

kiroii — Sun, 21 Oct 2007 09:22:43 +0000

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

By: kiroii

kiroii — Sun, 21 Oct 2007 09:20:46 +0000

erm fer the first part it the value of m and p given? if not i am truely clueless>.

By: Miss Loi

Miss Loi — Sun, 21 Oct 2007 05:37:03 +0000

Ok kiroii to cut a long story short: FOR PART 1, Distance to The Temple = log_b√(xy) Given that : log_b(xy³) = m and log_b(x³y²) = p, Express log_b√(xy) in terms of m and p. FOR PART 2, Distance to The Temple = 10^x Given that 2^2x+2 x 5^x-1 = 8^x x 5^2x, Find the value of 10^x. Note: The final answer for 10^x is numerical, there shouldn't be any more x present.

By: kiroii

kiroii — Sun, 21 Oct 2007 05:03:23 +0000

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-.-