求下面对数运算的详细过程

(a)
2logx -log(x+2) =log(2x-3)
定义域
x>0 and x>-2 and x>3/2
=> x>3/2
2logx -log(x+2) =log(2x-3)
log[x^2/(x+2)] =log(2x-3)
x^2/(x+2) =2x-3
x^2=(x+2)(2x-3)
x^2+x-6=0
(x+3)(x-2)=0
x=2 or -3(rej)
ie
x=2
(b)
log<6>(x+3) +log<6> (x+4) = 1
定义域
x>-3 and x>-4
ie
x>-3
log<6>(x+3) +log<6> (x+4) = 1
log<6>[(x+3)(x+4)] = 1
(x+3)(x+4) = 6
x^2+7x+6=0
(x+1)(x+6)=0
x=-1 or -6(rej)
ie
x=-1
(c)
2.5^(x-2) =100
5^(x-2) =50
x-2=log<5>50
= 2 + log<5>2
x=4 + log<5>2
(d)
2^(x+3)=6^(x-1)
2^(x+3)=2^(x-1).3^(x-1)
2^(x+3)-2^(x-1).3^(x-1) =0
2^(x-1) .[ 16-3^(x-1)] =0
16-3^(x-1)=0
3^(x-1)=16
x-1=4log<3>2
x=1+4log<3>2