What are the real solutions of the equation log(x+2) x + logx (x+2)=5/2 ?

We'll impose the constraints of existence of
logarithms:

x>0

x+2>0

x>-2

x
different from 1

x + 2 different from
1

x different from -1.

The
interval of admissible values for x is (0 ;
+infinite)-{1}.

Now, we'll solve the equation. We'll start
by changing the base with by the argument, in the 1st
term:

log(x+2) x = 1/logx
(x+2)

We'll note logx (x+2) =
t

We'll write the equation in
t:

1/t + t = 5/2

We'll
multiply by 2t both sides:

2 + 2t^2 =
5t

We'll move all terms to one
side:

2t^2 - 5t + 2 = 0

We'll
apply quadratic formula:

t1 = [5+sqrt(25 -
16)]/4

t1 = (5+3)/4

t1 =
2

t2 = (5-3)/4

t2 =
1/2

But logx (x+2) = t1 => logx (x+2) =
2

We'll take antilogarithms:

x
+ 2 = x^2

We'll subtract x+2 both
sides:

x^2 - x - 2 = 0

We'll
apply quadratic formula:

x1 =
[1+sqrt(1+8)]/2

x1 =
(1+3)/2

x1 = 2

x2 =
(1-3)/2

x2 = -1

We'll reject
the second value of x since it doesn't respect the constraints of existence of
logarithms.

Now, we'll put

But
logx (x+2) = t2 => logx (x+2) = 1/2

We'll take
antilogarithms:

x + 2 =
sqrtx

We'll raise to square both
sides:

(x+2)^2 = x

We'll
expand the square:

x^2 + 4x + 4 =
x

x^2 + 3x + 4 = 0

We'll apply
quadratic formula:

x1 =
[-3+sqrt(9-16)]/2

Since sqrt-7 is undefined, the equation
x^2 + 3x + 4 = 0 has no real solutions.

The
equation will have only one valid solution: x =
2.