Question

solve for x:
1. log6(x+3)+log6(x+4) = 1 "log base 6"

Answer 1

hmm why divide?

Answer 2

log M + log N = Log MN right?

Answer 3

\[\log_6[(x+3)(x+4)]=1\]
\[(x+3)(x+4)=6\]
\[x^2+4x+3x+12=6\]
\[x^2+7x+12-6=0\]
\[x^2+7x+6=0\]
\[(x+6)(x+1)=0\]
\[x=-6,x=-1\]
but only x=-1 works

Answer 4

-6 doesnt work because it makes the log negative right?

Answer 5

yes
-6+4 is negative
-6+3 is negative