Solve for x in the equation:log_10 x+log _10 (x+2)=2

Answer 1

Using the logarithmic property $\log_a b + \log_a c = \log_a (b \cdot c)$, we can rewrite the equation as: $\log_{10} (x \cdot (x+2)) = 2$.

Since $\log_a b = c$ is equivalent to $a^c = b$, we have $10^2 = x \cdot (x+2)$.

Simplifying, we get $100 = x^2 + 2x$.

Moving all terms to one side, we have $x^2 + 2x - 100 = 0$.

Factoring or using the quadratic formula to solve this quadratic equation, we find that $x = -10$ or $x = 8$.

However, since $\log_{10} x$ requires $x$ to be positive, we discard the solution $x = -10$.

Therefore, the solution to the equation $\log_{10} x + \log_{10} (x+2) = 2$ is $x = 8$.