Q: The function is $$k(\\theta)=3\\sin(2\\theta)$$ Find all values of $$\\theta$$ in $$[0,2\\pi)$$ such that $$k(\\theta)=-\\frac{3}{2}$$.

Set:$$3\sin(2\theta)=-\frac{3}{2}.$$Divide by $$3.$$$$\sin(2\theta)=-\frac{1}{2}.$$So,$$2\theta=\frac{7\pi}{6},\frac{11\pi}{6}.$$Divide by $$2.$$$$\theta=\frac{7\pi}{12},\frac{11\pi}{12}.$$Final answer: $$\frac{7\pi}{12},\frac{11\pi}{12}$$

Question 1

Consider the functions $$g(x)=8^{x-1}$$ and $$h(x)=2^{3x+5}$$ Find the $$x$$-coordinate of their intersection.

Accepted Answer

Rewrite with base $$2.$$$$8=2^3.$$So:$$8^{x-1}=2^{3(x-1)}.$$Thus the equation becomes:$$2^{3(x-1)}=2^{3x+5}.$$Simplify the exponents:$$2^{3x-3}=2^{3x+5}.$$Equate exponents:$$3x-3=3x+5.$$$$-3=5.$$This is a contradiction, so there is no solution from direct exponent comparison.Check the equation numerically.Solve:$$8^{x-1}=2^{3x+5}.$$Try $$x=-2.$$Left side:$$8^{-3}=\frac{1}{512}.$$Right side:$$2^{3(-2)+5}=2^{-6+5}=2^{-1}=\frac{1}{2}.$$These are not equal, so there is no real solution.Final answer: $$	ext{no solution}$$

Question 2

The function is given by $$f(x)=\log_3(2x-1)$$ Find the value of $$x$$ such that $$f(x)=4$$.

Accepted Answer

Set:$$\log_3(2x-1)=4.$$Rewrite in exponential form:$$2x-1=3^4.$$$$2x-1=81.$$$$2x=82.$$$$x=41.$$Final answer: $$41$$

Question 3

The function is $$k(\theta)=3\sin(2\theta)$$ Find all values of $$\theta$$ in $$[0,2\pi)$$ such that $$k(\theta)=-\frac{3}{2}$$.

Accepted Answer

Set:$$3\sin(2	heta)=-\frac{3}{2}.$$Divide by $$3.$$$$\sin(2	heta)=-\frac{1}{2}.$$So,$$2	heta=\frac{7\pi}{6},\frac{11\pi}{6}.$$Divide by $$2.$$$$	heta=\frac{7\pi}{12},\frac{11\pi}{12}.$$Final answer: $$\frac{7\pi}{12},\frac{11\pi}{12}$$

Question 4

Which expression is equivalent to $$\log_5(x^8y^2)$$?

Accepted Answer

Use logarithm rules.$$\log(ab)=\log a+\log b$$$$\log(a^n)=n\log a$$So:$$\log_5(x^8y^2)=8\log_5 x+2\log_5 y.$$Final answer: $$8\log_5 x+2\log_5 y$$

Mini Exam 1

Frequently Asked Questions