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# Tutor profile: Rishabh S.

Inactive
Rishabh S.
Math Tutor
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## Questions

### Subject:Pre-Calculus

TutorMe
Question:

Expand the logarithmic expression: $$\log_{a}(x\sqrt{x^{2}+1})$$

Inactive
Rishabh S.
Answer:

$$=> \log_{a}(x\sqrt{x^{2}+1})$$ $$=> \log_{a}(x)+ \log_{a}(\sqrt{x^{2}+1})$$ Using the rule: $$\log_{a}(M*N)=\log_{a}(M)+\log_{a}(N)$$ $$=> \log_{a}(x)+ \log_{a}({x^{2}+1})^{\frac{1}{2}}$$ $$=> \log_{a}(x)+\frac{1}{2} \log_{a}({x^{2}+1})$$ Using the rule: $$\log_{a}(M)^{r}=r\log_{a}(M)$$

### Subject:Trigonometry

TutorMe
Question:

Solve the equation: $$2sin\theta + \sqrt{3} = 0$$ ; $$[0,2\pi)$$

Inactive
Rishabh S.
Answer:

$$=> 2sin\theta + \sqrt{3} = 0$$ ; $$[0,2\pi)$$ $$2sin\theta =- \sqrt{3}$$ $$sin\theta = \frac{-\sqrt{3}}{2}$$ In the interval $$[0,2\pi)$$, $$sin\theta = \frac{-\sqrt{3}}{2}$$ for only two angles, specifically for $$\theta=\frac{4\pi}{3}$$ and $$\theta=\frac{5\pi}{3}$$ Therefore, the solution set is {$$\frac{4\pi}{3} , \frac{5\pi}{3}$$}.

### Subject:Algebra

TutorMe
Question:

Solve the equation: $$|2x-1|+1=6$$

Inactive
Rishabh S.
Answer:

$$=> |2x-1|+1=6$$ $$=> |2x-1|=6-1$$ $$=> |2x-1|=5$$ $$=> 2x-1=5$$ or $$2x-1=-5$$ $$=> 2x=5+1$$ or $$2x=-5+1$$ $$=> 2x=6$$ or $$2x=-4$$ $$=> x=\frac{6}{2}$$ or $$x=\frac{-4}{2}$$ $$=> x=3$$ or $$x=-2$$ Therefore, the solution set is {-2,3}.

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