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