identify the error in the student solution shown below. find the correct answer.\n2ln(x)=ln(3x)-ln(9)-2ln(3)\…

Answer:

The error is in the step $\ln(x^{2})=\ln(3x/0)$. When we have $\ln(x^{2})=\ln(3x)-0$, we should just have $\ln(x^{2})=\ln(3x)$ since subtracting 0 from a quantity does not change it and there is no division operation involved. The correct solution: Since $\ln(x^{2})=\ln(3x)$, then $x^{2}=3x$ (because the natural - logarithm function $y = \ln(u)$ is one - to - one for $u>0$). Rearranging gives $x^{2}-3x = 0$, factoring out an $x$ we get $x(x - 3)=0$. So $x=0$ or $x = 3$. But when $x = 0$, $\ln(x)$ and $\ln(3x)$ are undefined in the original equation. So the solution is $x = 3$.

Explanation:

Step1: Identify the error

The step $\ln(x^{2})=\ln(3x)-0$ was wrongly transformed to $\ln(x^{2})=\ln(3x/0)$. Subtracting 0 doesn't imply division by 0.

Step2: Correct the equation

Since $\ln(x^{2})=\ln(3x)-0$, we have $\ln(x^{2})=\ln(3x)$.

Step3: Use one - to - one property of $\ln$

If $\ln(a)=\ln(b)$ and the domain of $\ln$ is $a>0,b>0$, then $a = b$. So $x^{2}=3x$.

Step4: Rearrange and factor

$x^{2}-3x=0$, factor to $x(x - 3)=0$.

Step5: Solve for $x$

We get $x=0$ or $x = 3$.

Step6: Check domain

In the original equation $2\ln(x)=\ln(3x)-[\ln(9)-2\ln(3)]$, $x>0$ and $3x>0$. So $x = 0$ is an extraneous solution. The solution is $x = 3$.