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

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### 2 Answers

We have to find the real solutions of log(x+2) x + log(x) (x+2) = 5/2.

We use the relation log (a) b = 1/log (b) a and log a + log b = log a*b

log(x+2) x + log(x) (x+2) = 5/2

=> 1/ log (x) (x+2) + log(x) (x+2) = 5/2

let log(x) (x + 2) = y

=> 1/y + y = 5/2

=> 1 + y^2 = 5y/2

=> 2 + 2y^2 = 5y

=> 2y^2 - 5y + 2 = 0

=> 2y^2 - 4y - y + 2 = 0

=> 2y( y - 2) - 1(y - 2) = 0

=> (2y - 1)( y - 2) = 0

=> y = 1/2 and y = 2

log(x) (x + 2) = 2

=> x + 2 = x^2

=> x^2 - x - 2 = 0

=> x^2 - 2x + x - 2 = 0

=> x(x - 2)+ 1(x - 2) = 0

=> (x + 1)(x - 2) = 0

=> x = -1 and x = 2

We cannot have x = -1 as log of a negative number is not defined.

log(x) (x + 2) = 1/2 yields complex values of x which are not defined.

**The value is x = 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.**