math Log數~ 急

1)logx + log(1+y)^2 = log 9y/x
log [x(1+y)^2] = log 9y/x.............(logM + logN = logMN)
x(1+y)^2 = 9y/x
x^2 = 9y / (1+y)^2
x = (+ or - ) [3y^(1/2)] / (1+y)
2) 3(5^x+3)=4(72^x-1)
(3*5^3) (5^x) = (4/72) (72^x)
6750 = (72^x) / (5^x)
6750 = (72/5)^x
log 6750 = log (72/5)^x
log 6750 = x log (72/5)
x = (log 6750) / log (72/5)
x = 3.305790544 = 3.31 (correct to 2 decimal places.)


2009-09-18 19:46:34 補充：
Corrections :

For Q1 , x = - [3y^(1/2)] / (1+y) is rejected , since x cannot be negative for logx