# Tutor profile: Olivia T.

## Questions

### Subject: Pre-Calculus

Solve the equation $$ln(x)+ln(x+2)=ln(x+6)$$.

We know that when adding two logarithms of the same base, here the base is $$e$$, we can combine the two logarithms into one by multiplying the inside of the logarithm. So, $$ln(x)+ln(x+2)=ln(x(x+2))=ln(x^2+2x)$$. Therefore, $$ln(x^2+2x)=ln(x+6)$$. Now, we can raise $$e$$ to both sides of the equation. So, $$e^{ln(x^2+2x)}=e^{ln(x+6)}$$. Since $$e$$ and $$ln$$ are inverses, we see that $$x^2+2x=x+6$$. Therefore, $$x^2+x-6=0$$. By factoring, we find that $$x=-3$$ or $$x=2$$. By checking our original equation, we see that $$x=-3$$ is not in our domain. So, the only solution is $$x=2$$.

### Subject: Calculus

Find the derivative of the function $$f(x)= 4e^{4x}+3x^2+9$$.

We first notice that we will be using the chain rule, power rule, constant rule, addition rule, constant multiplier rule, and the exponential rule. First, we know that $$\frac{d}{dx}4e^{4x} = 4(e^{4x})(\frac{d}{dx}4x)=4e^{4x}(4)=16e^{4x}.$$ Now, $$\frac{d}{dx}3x^2= 3(2x)=6x.$$ And, $$\frac{d}{dx}9=0.$$ Since the derivative of a sum, is the sum of the derivatives, we know that $$\frac{d}{dx}(4e^{4x}+3x^2+9) =16e^{4x}+6x+0=16e^{4x}+6x$$

### Subject: Algebra

Factor the equation $$x^2+4x-5$$.

We want to first begin by seeing whether or not we can factor using the rules that we know. So, we want to see if there are two numbers that multiply to get -5, but add to get 4. We notice that there are no numbers. So we can use the quadratic formula. The quadratic formula is $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$. In this case, $$a=1, b=4, c=-5.$$ So, $$x=\frac{-4\pm\sqrt{4^2-4(1)(-5)}}{2(1)}=\frac{-4\pm\sqrt{16+20}}{2}=\frac{-4\pm\sqrt{36}}{2}=\frac{-4\pm6}{2}.$$ Therefore, $$x=-5$$ and $$x=1$$.

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