中4 math

Q1

logx + log(x-3) = 1
log(x)(x-3) = 1
x² - 3x = 10
x² - 3x -10 = 0
(x - 5)(x + 2) = 0
x = 5 or x = -2

x=-2 should be rejected since log(-2) is not possible, therefore x = 5

Q2

log(2x-10) - 2logx= -1
log( (2x-10) / x² ) = -1

(2x-10) / x² = 1 / 10
20x - 100 = x²
x² -20x + 100 =0
(x-10)² = 0
x=10

logx + log(x-3)=1
log(x)(x-3)=log 10
x^2-3x-10=0
(x-5)(x+2)=0
x=5 or x=-2

log(2x-10) - 2logx= -1
log((2x-10)/x^2)=log 0.1
2x-10/x^2=1/10
x^2-20x+100=0
(x-10)^2=0
x=10

Q.1

logx + log(x-3)=1
log [x(x+3)] = log 10
x(x+3) = 10
x²+3x-10 = 0
(x-5)(x+2) = 0
x= 5 or x= -2(reject)
so x=5

Q.2

log(2x-10) - 2logx= -1
log(2x-10) - log x² = log (0.1)
log[(2x-10) / x²] = log(0.1)
(2x-10)/x² = 0.1
2x-10 = 0.1x²
20x-100=x²
x²-20x+100 = 0
(x-10)²=0
x=10