Here x must equal positive 1. Nice problem solving if you figured this out. We
start solving this problem by multiplying x times x plus 2. Since we're adding
two logs with the same base, we can just multiply these arguments together. So
we'll have log base 4 of x times x, which is x squared, and x times 2, which is
plus 2x. The log base 4 of 3 stays on the right-hand side of the equation. Now
that we have a logarithm on one side of the equation here and a logarithm of
something else here, we can set these two things equal to one another. We
want to factor this quadratic, but first we need to subtract 3 from both sides,
to set it equal to zero. Now factoring, we get x plus 3 times x minus 1. We set
each of these factors equal to 0 so x can equal negative 3 or x can equal
positive 1. Now keep in mind that we want these arguments to be positive so
that way we can take a log of them. So if x were equal to negative 3, I would
have a negative number here. I can't take the log base 4 of negative 3. That's
not possible. So this solution is out. This means the only solution is x equals
positive 1.
