what is the true solution to ln 20+ln 5 = 2 ln x?\no x = 5\no x = 10\no x = 50\no x = 100

Explanation:

Step1: Apply log - addition rule

Using the rule $\ln a+\ln b=\ln(ab)$, we have $\ln(20\times5)=\ln100$. The left - hand side of the equation $\ln 20+\ln 5$ becomes $\ln100$. The right - hand side, using the rule $n\ln a=\ln(a^{n})$, $2\ln x=\ln(x^{2})$. So the equation is $\ln100 = \ln(x^{2})$.

Step2: Solve for x

Since the natural logarithm function $y = \ln u$ is one - to - one, if $\ln100=\ln(x^{2})$, then $x^{2}=100$. Taking the square root of both sides, we get $x=\pm10$. But since the domain of $\ln x$ requires $x>0$, we discard $x = - 10$. So $x = 10$.

Answer:

B. $x = 10$