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

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

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