A. Maths Questions - surds and logarithms replied by weewee @ Sat, 05 Jul 2008 01:10:13 +0800

weewee — Sat, 05 Jul 2008 01:10:13 +0800

thanks.

A. Maths Questions - surds and logarithms replied by secretliker @ Sun, 29 Jun 2008 02:35:04 +0800

secretliker — Sun, 29 Jun 2008 02:35:04 +0800

Q1) If a^(2x-1) = b^(1-3y) and a^(3x-1) = b^(2y-2), show that 13xy = 7x + 5y -3.

(2x-1)(2y-2) = (1-3y)(3x-1)

4xy-2y-4x+2 = 3x-9xy-1+3y

13xy = 7x + 5y - 3

Q2) Given that log5 (x) = 4 logx (5), calculate the possible values of x.

log5(x) / log5(5) = 4 log5(5) / log5(x)

[ log5(x) ] ^2 = 4 [ log5(5) ] ^2

[ log5(x) ] = ±(2)

x=5^(-2) or 5^2

x=1/25 or 25

weewee — Sat, 28 Jun 2008 22:56:30 +0800

omg thanks! :)

jiaxing2 — Sat, 28 Jun 2008 22:46:49 +0800

for qn 1

multiply log base 10 on both sides of both eqn

u will get (2x-1)/(1-3y) = lg b / lg a----(1)

(3x-1)/(2y-2) = lg b / lg a----(2)

(1) = (2)

thus 4xy + 2 - 4x- 2y = 3x + 3y- 1- 9xy

Hence 13xy = 5y + 7x - 3

Qn 2)

Re-express the eqn with lg base 10

lg x / lg 5 = lg 625 / lg x (5^4=625)

(lg x)^2 = lg 5 ( lg 625 )

solve for values of x using ur own calculator

weewee — Sat, 28 Jun 2008 22:01:35 +0800

I need help on these two questions.

Q1) If a^(2x-1) = b^(1-3y) and a^(3x-1) = b^(2y-2), show that 13xy = 7x + 5y -3.

Q2) Given that log5 (x) = 4 logx (5), calculate the possible values of x.

For question 2, the first term is log with base 5 and the second term is log with base x.