Question

log6(x+3)+log6(x+4)=1

Answer 1

x = -1.

Answer 2

\[\log_6 (x+3) + \log_6 ( x+4) = 1\]When you add logs of the same base, you can multiply the thing inside the log to combine them.\[\log_6 ((x+3)(x+4))= 1\]

Answer 3

\[\log_6 (x^2 + 7x + 12) = 1\]Now do 6^both sides to cancel out the log\[6^{\log_6 (x^2 + 7x + 12)} = 6^{1}\]\[x^2 + 7x + 12 = 6\]\[x^2 + 7x + 6 = 0\]\[(x + 1)(x+6) = 0\]\[x = -1, - 6\]Checking the solutions, -6 is not possible, so the only answer is -1.