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.
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
omg thanks! :)
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
thanks.