Enable contrast version

Tutor profile: Rajinda W.

Inactive
Rajinda W.
Doctoral Student in Applied Mathematics
Tutor Satisfaction Guarantee

Questions

Subject: Pre-Calculus

TutorMe
Question:

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

Inactive
Rajinda W.
Answer:

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

TutorMe
Question:

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

Inactive
Rajinda W.
Answer:

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

TutorMe
Question:

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

Inactive
Rajinda W.
Answer:

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}

Contact tutor

Send a message explaining your
needs and Rajinda will reply soon.
Contact Rajinda

Request lesson

Ready now? Request a lesson.
Start Lesson

FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.
BEST IN CLASS SINCE 2015
TutorMe homepage
Made in California by Zovio
© 2013 - 2021 TutorMe, LLC
High Contrast Mode
On
Off