But note that x = -6 cannot be used as a solution because it is undefined for the terms log_6 (x + 3) and log_6 (x + 4). That is, if x = -6, we get log_6 (-3) and log_6 (-2) and these values are undefined.

log₆(x+3) + log₆(x+4) = 1 : x >-3 log₆[(x+3)(x+4)] = 1 (x+3)(x+4) = 6 x² + 7x + 6 = 0 (x+6)(x+1) = 0 x = -6, -1 Discard the x = -6 result as x must be greater than -3 for the equation to be real valued.

## Answers

log_6 (x + 3) + log_6 (x + 4) = 1

Note these rules:

1) log (ab) = log(a) + log(b)

2) log_a (a) = 1

So we can rewrite the equation as:

log_6 [(x + 3)(x + 4)] = log_6 (6)

We can take the anti-log of both sides to convert this equation to one not involving logarithms:

(x + 3) (x + 4) = 6

Expand brackets, simplify and factorize:

x^2 + 4x + 3x + 12 = 6

x^2 + 7x + 12 = 6

x^2 + 7x + 6 = 0

x^2 + 6x + x + 6 = 0

x(x + 6) + 1(x + 6) = 0

(x + 6)(x + 1) = 0

x = -6 or x = -1

But note that x = -6 cannot be used as a solution because it is undefined for the terms log_6 (x + 3) and log_6 (x + 4).

That is, if x = -6, we get log_6 (-3) and log_6 (-2) and these values are undefined.

So the solution is x = -1

log₆(x + 3) + log₆(x + 4) = 1

given log(a) + log(b) = log(ab)

∴ log₆((x + 3)(x + 4)) = 1

∴ 6^log₆((x + 3)(x + 4)) = 6¹

given a^log_a(n) = n

∴ (x + 3)(x + 4) = 6

∴ x² + 7x + 12 = 6

∴ x² + 7x + 6 = 0

∴ (x + 1)(x + 6) = 0

∴ x = -1 and x = -6

given x = -6 is outside of domain of the original question

∴ x = -1

log₆(x+3) + log₆(x+4) = 1 : x >-3

log₆[(x+3)(x+4)] = 1

(x+3)(x+4) = 6

x² + 7x + 6 = 0

(x+6)(x+1) = 0

x = -6, -1

Discard the x = -6 result as x must be greater than -3 for the equation to be real valued.

x = -1

log(base 6)(x + 3) + log(base 6)(x + 4) = 1

log(base 6)((x + 3)(x + 4)) = 1

log(base 6)(x^2 + 7x + 12) = 1

6^1 = x^2 + 7x + 12

0 = x^2 + 7x + 6

(x + 6)(x + 1) = 0

x = -6 or x = -1

Discard x = -6 because it leads to imaginary numbers , leaving just

x = -1

(x+3)(x+4)=6 , solve for x , with x >-3