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# Tutor profile: Amy N.

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Amy N.
Experienced tutor/teacher with background in chemical engineering & mathematics
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## Questions

### Subject:Pre-Calculus

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Question:

Solve the equation $$16^{3x} = 8^{5x + 1}$$ for $$x$$.

Inactive
Amy N.

We first need both sides of the equation to be expressed in terms of the same base. To that end, recall that $$8 = 2^{3}$$ and $$16 = 2^{4}$$. Thus, we can rewrite the equation as $$(2^{4})^{3x} = (2^{3})^{5x + 1}$$, which simplifies to $$2^{4 \times 3x} = 2^{3 \times (5x + 1)}$$, which in turn simplifies to $$2^{12x} = 2^{15x + 3}$$. Taking the base-2 logarithm of both sides of this equation yields 12x = 15x + 3. This equation simplifies to 3x = 3, which yields x = 1.

### Subject:Calculus

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Question:

Evaluate the integral $$\int^{1}_{-1} x^{3} dx$$ without performing integration.

Inactive
Amy N.

Notice that $$x^{3}$$ is an odd function: $$(-x)^{3} = -x^{3}$$. As a result, it is symmetric about the origin, which happens to be the midpoint of the interval of integration, $$[-1,1]$$. Therefore, $$\int^{0}_{-1} x^{3} dx = - \int^{1}_{0} x^{3} dx$$. Since $$\int^{0}_{-1} x^{3} dx + \int^{1}_{0} x^{3} dx = \int^{1}_{-1} x^{3} dx$$, we find that $$\int^{1}_{-1} x^{3} dx = - \int^{1}_{0} x^{3} dx + \int^{1}_{0} x^{3} dx = 0$$.

### Subject:Algebra

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Question:

Currently, Adam is 2 years older than Max. In 3 years, Max's age will be 5 less than twice Adam's age. How old are Adam and Max right now?

Inactive
Amy N.

Let A = Adam's current age and M = Max's current age. Because Adam is currently 2 years older than Max, we have one equation given by A = M + 2 (eqn. 1). Then because in 3 years Max's age will be 5 less than 2 times Adam's age, we have another equation given by M + 3 = 2 (A + 3) - 5 (eqn. 2). Substituting eqn. 1 into eqn. 2 yields M + 3 = 2 (M + 2 + 3) - 5 = 2M - 5, which simplifies to M = 8. Plugging this result into eqn. 1 yields A = 8 + 2 = 10. Thus, Adam is currently 10 years old, and Max is 8 years old.

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