log(x)+log(x + 1)=3\nx =

Explanation:

Step1: Use log - product rule

Using the rule $\log(a)+\log(b)=\log(ab)$, we rewrite the left - hand side of the equation: $\log(x(x + 1))=3$.

Step2: Convert from logarithmic to exponential form

If $\log(x(x + 1))=3$, and assuming base 10 logarithm, we have $x(x + 1)=10^{3}$ since if $\log_{a}y=x$, then $y=a^{x}$. So $x(x + 1)=1000$.

Step3: Expand and form a quadratic equation

Expand $x(x + 1)$ to get $x^{2}+x - 1000=0$.

Step4: Use the quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 1$, $b = 1$, and $c=-1000$. So $x=\frac{-1\pm\sqrt{1^{2}-4\times1\times(-1000)}}{2\times1}=\frac{-1\pm\sqrt{1 + 4000}}{2}=\frac{-1\pm\sqrt{4001}}{2}$. Since $x>0$ (because $\log(x)$ is only defined for $x>0$), we take the positive root.

Answer:

$x=\frac{-1+\sqrt{4001}}{2}$