Solve the following equations ln x ln 3 lnx 2 logx 3 1

Solve the following equations. ln x = ln 3 - ln(x - 2). log(x + 3) = 1 - log(x - 2). log(x - 1) = log(2/x) + log(3x - 5).

1) 4^(2x +3) =16.2^(x-2)

RHS 16.2^(x-2) = 2^4^(x-2) = 2^(x+2)

LHS 4^(2x +3) =2^2(2x +3) = 2^(4x+6)

2^(x+2) = 2^(4x+6)

when bases are same we can equate the exponents:

4x +6 = x+2

3x = 4

x = 4/3

2) lnx = ln3 - ln(x -2)

use the property : lnA - lnB = lnA/B

RHS : ln3 - ln(x -2) = ln(3/x-2)

lnx = ln(3/(x-2) )

equate the argument inside the log:

x = 3/(x -2)

x(x-2) =3

x^2 -2x -3 =0

factorise : x^2 -3x +x -3 =0

x(x -3) +1(x-3) =0

(x +1)(x -3) =0

x = -1 , 3

3) log(x +3) = 1- log(x -2)

log(x +3) +log(x -2) = 1

log[(x +3)(x-2)] =1

(x+3)(x-2) = 10^1

x^2 +x -6 =10

x^2 + x -16 =0

Roots of quadratic equation cab be found using quadratic formula:

x = (-b +/- sqrt(b^2 -4ac) )/2a

= ( -1 +/ sqrt(65)/2

= -1/2 +/- sqrt(65)/2