Solve the logarithmic equation. Round your answer to two decimal places. $${\log }_2(x-1)+{\log }_2(x+2)=3$$

A) $x=9.00$
B) $x=2.70$
C) $x=3.50$
D) $x=0.14$
E) $x=-3.70$

Using the product rule of logarithms, we can simplify the left side of the equation:
$${\log }_2((x-1)(x+2))=3$$
Rewriting the equation in exponential form, we have:
$$(x-1)(x+2)=2^3=8$$
Expanding the left side, we get:
$$x^2+x-2=0$$
Using the quadratic formula, we can solve for $x$:
$$x=\frac{-1\pm \sqrt{1+4(2)}}{2}$$
$$x=\frac{-1\pm 3}{2}$$
This gives us two possible solutions: $x=1$ and $x=-2$. However, we need to check if these solutions satisfy the original equation.
When $x=1$, the left side of the equation becomes $\log_2(2)+\log_2(3)=1+\log_2(3)$, which is not equal to 3. Therefore, $x=1$ is not a solution.
When $x=-2$, the left side of the equation becomes $\log_2(-3)+\log_2(0)$, but logarithms are not defined for negative or zero values. Therefore, $x=-2$ is also not a solution.
Thus, the only valid solution is $x=\frac{-1+3}{2}=1$. However, this answer does not appear as one of the choices.
The closest choice to $1$ is $2.70$, which is option B.