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# Tutor profile: Rajinda W.

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Rajinda W.
Doctoral Student in Applied Mathematics
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## Questions

### Subject:Pre-Calculus

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Question:

Solve the equation $$\log(3x) = \log 2 + \log(x + 2)$$.

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Rajinda W.

First we use the property $$\log A+\log B=\log (AB).$$ Therefore, \begin{equation} \log(3x) = \log[2(x + 2)] \implies \log(3x) = \log[2x+4]. \end{equation} Next, we use the property $$a^{\log_a x}=x$$ to remove the log from both sides of the equation. We can write both sides as the exponent of the base of the log, which is 10 in this case. \begin{equation} 10^{\log(3x)} = 10^{\log[2x+4]}\implies 3x=2x+4. \end{equation} When we solve $$3x=2x+4$$, we get that $$x=4.$$ When solving logarithmic equations, always check the solutions you get at the end to verify if they are in elements in the domains of the logarithm. Recall that the domain of $$\log x$$ is $$(0,\infty).$$ Therefore, we cannot end up with anything negative inside any of the logarithms in the equation. If we substitute $$x=4$$ into the initial equation, we can easily see that we don't get any negative numbers or 0: \begin{equation} \log(3\cdot4) = \log 2 + \log(4 + 2). \end{equation} Therefore, $$x=4$$ is the only solution.

### Subject:Trigonometry

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Question:

Find the exact value of $$\sin \left(\frac{\pi}{12}\right)$$.

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Rajinda W.

The trick here is to use the difference formula  \begin{equation} \sin (A-B)=\sin A \cos B-\cos A \sin B. \end{equation} Unfortunately, $$\frac{\pi}{12}$$ is not a special angle on the unit circle. Therefore, we split $$\frac{\pi}{12}$$ as the difference of two angles, that are special angles on the unit circle. Notice that \begin{equation} \sin \left(\frac{\pi}{12}\right)=\sin \left(\frac{3\pi}{12}-\frac{2\pi}{12}\right)=\sin \left(\frac{\pi}{4}-\frac{\pi}{6}\right). \end{equation} Now apply the difference formula to $$\sin \left(\frac{\pi}{4}-\frac{\pi}{6}\right):$$ \begin{equation} \sin \left(\frac{\pi}{4}-\frac{\pi}{6}\right)=\sin \frac{\pi}{4} \cos \frac{\pi}{6}-\cos \frac{\pi}{4} \sin \frac{\pi}{6}=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}. \end{equation}

### Subject:Calculus

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Question:

Find the derivative of the function $$f(x)=x^x$$.

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Rajinda W.

This question is tricky and you need to understand that you cannot use the power rule here. In particular, $$\frac{d}{dx}(x^x)\neq x\cdot x^{x-1}.$$ The reason is, the power rule only works when the exponent is a constant, not a variable! Therefore, we need a different game plan to differentiate this function. Let's start with the given function: \begin{equation} y=x^x. \end{equation} Next, take the natural log of both sides: \begin{equation} \ln y=x\ln x. \end{equation} Now, we differentiate the above equations with respect to $$x$$. Keep in mind that $$y$$ is a function of $$x$$. Therefore, \begin{equation} \frac{1}{y}\cdot\frac{dy}{dx}=x\cdot \frac{1}{x}+1\cdot\ln x.=1+\ln x \end{equation} Multiplying both sides by $$y$$ yields \begin{equation} \frac{dy}{dx}=y(1+\ln x) \end{equation} Finally, setting $$y=x^x$$ gives us the final answer: \begin{equation} \frac{dy}{dx}=x^x(1+\ln x). \end{equation}

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