# Solve the following equation: log (x + 3) + log (x^2 + 1) – log ( 5x + 3) = log 1

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We have to solve the equation log (x + 3) + log (x^2 + 1) – log (5x + 3) = log 1

First simplify using the relation: log a + log b = log (a*b) and log a - log b = log (a/b)

log (x + 3) + log (x^2 + 1) – log (5x + 3) = log 1

=> log [(x + 3) (x^2 +1)/ (5x +3)] = log 1

taking the anti log of both the sides

=> [(x + 3) (x^2 +1)/ (5x +3)] = 1

=> (x + 3) (x^2 +1) = (5x +3)

=> x^3 + 3x^2 + x +3 = 5x +3

=> x^3 + 3x^2 – 4x = 0

=> x(x^2 + 3x – 4) = 0

=> x [x^2 + 4x - x – 4] = 0

=> x [x(x + 4) – 1(x +4)] = 0

=> x (x -1) (x + 4) = 0

So x can be 0, 1 and -4.

We eliminate x = -4 as log (5x +3) and log (x +3) are not defined for x = -4.

**Therefore the required values of x are 0 and 1.**