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High School Math : Functions and Graphs

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Example Questions

Example Question #21 : Pre Calculus

Simplify the following polynomial function:

$ab^{-2}(a^{2}b + a^{-1}b^{3}-a^{3}b^{-1})$

Possible Answers:

$\frac{a^{3}}{b}-b+\frac{a^{4}}{b^{3}}$

$\frac{a^{3}}{b}-\frac{a^{4}}{b^{3}}$

$\frac{a^{3}}{b}+b^{2}-\frac{a^{4}}{b^{3}}$

$\frac{a^{3}}{b}+b-\frac{a^{4}}{b^{3}}$

$\frac{a^{3}}{b}+\frac{a^{4}}{b^{3}}$

Correct answer:

$\frac{a^{3}}{b}+b-\frac{a^{4}}{b^{3}}$

Explanation:

First, multiply the outside term with each term within the parentheses:

$ab^{-2}(a^{2}b + a^{-1}b^{3}-a^{3}b^{-1})$

$a^{3}b^{-1} + b-a^{4}b^{-3}$

Rearranging the polynomial into fractional form, we get:

$\frac{a^{3}}{b}+b-\frac{a^{4}}{b^{3}}$

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Example Question #21 : Pre Calculus

You are given that $\log_{10} x = A$ and $\log_{10} y = B$ .

Which of the following is equal to $100^{2A-B}$ ?

Possible Answers:

$\frac{2x}{y}$

$\frac{x^{4}}{y^{2}}$

$\frac{y^{2}}{x^{4}}$

$\frac{x}{\sqrt{y}}$

$x^{4}y^{2}$

Correct answer:

$\frac{x^{4}}{y^{2}}$

Explanation:

Since $\log_{10} x = A$ and $\log_{10} y = B$ , it follows that $10^{A} = x$ and $10^{B} = y$

$100^{2A-B} = \left( 10^{2} \right ) ^{2A-B} =10^{2 (2A-B)} =10^{4A-2B}=\frac{10^{4A}}{10^{2B}} =\frac{\left ( 10^{A}\right )^4}{\left (10^{B}\right )^2}=\frac{x^4}{y^2}$

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Example Question #1 : Solving Logarithms

$\textup{Solve for }x\textup{: }\,\log_{2}32=x$

Possible Answers:

Correct answer:

Explanation:

$\textup{This can be re-written as an exponent problem: } 2^{x}=32$

$2\times2\times2\times2\times2=32\:\textup{ so }32=2^{5}\:\:\:x=5$

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Example Question #2 : Solving Logarithms

What is $log_{2}(8)$ ?

Possible Answers:

$\frac{1}{8}$

Correct answer:

Explanation:

Recall that by definition a logarithm is the inverse of the exponential function. Thus, our logarithm corresponds to the value of in the equation:

$2^{x} = 8$ .

We know that $2^{3} = 8$ and thus our answer is .

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Example Question #23 : Pre Calculus

Solve for : $\log_{2}\left ( x+4 \right ) + \log_{2}\left ( x+5 \right ) = \log_{2} 6$

Possible Answers:

The correct solution set is not included among the other choices.

$x = 4 \textrm{ or }x = 128$

$x = 1 \textrm{ or }x = \log_{2}7$

$x = 2 \textrm{ or }x = 7$

$x = -7 \textrm{ or }x = -2$

Correct answer:

The correct solution set is not included among the other choices.

Explanation:

$\log_{2}\left ( x+4 \right ) + \log_{2}\left ( x+5 \right ) = \log_{2} 6$

$\log_{2} \left[ \left ( x+4 \right )\left ( x+5 \right ) \right ] = \log_{2} 6$

$2^{ \log_{2} \left[ \left ( x+4 \right )\left ( x+5 \right ) \right ] }=2^{ \log_{2} 6}$

(x + 4)(x + 5) = 6

FOIL: $x^{2} + 4x + 5x + 4 \cdot 5 = 6$

$x^{2} + 9x + 20 = 6$

$x^{2} + 9x + 20 - 6 = 6- 6$

$x^{2} + 9x + 14 = 0$

(x + 2) (x + 7) = 0

$x + 2 = 0 \textrm{ or }x + 7 = 0$

$x = -2 \textrm{ or } x = -7$

These are our possible solutions. However, we need to test them.

$\log_{2}\left ( x+4 \right ) + \log_{2}\left ( x+5 \right ) = \log_{2} 6$

$\log_{2}\left ( -2+4 \right ) + \log_{2}\left ( -2+5 \right ) = \log_{2} 6$

$\log_{2}2 + \log_{2}3 = \log_{2} 6$

$\log_{2}2 = 1$

$\log_{2}3 = \frac{\ln 3}{\ln 2} \approx \frac{1.10}{0.69} \approx 1.59$

$\log_{2}6 = \frac{\ln 6}{\ln 2} \approx \frac{1.79}{0.69} \approx 2.59$

The equation becomes 1 + 1.59 = 2.59 . This is true, so x = -2 is a solution.

x = -7 :

$\log_{2}\left ( x+4 \right ) + \log_{2}\left ( x+5 \right ) = \log_{2} 6$

$\log_{2}\left ( -7+4 \right ) + \log_{2}\left ( -7+5 \right ) = \log_{2} 6$

$\log_{2}\left (-3 \right ) + \log_{2}\left (-2 \right ) = \log_{2} 6$

However, negative numbers do not have logarithms, so this equation is meaningless. x = -7 is not a solution, and x = -2 is the one and only solution. Since this is not one of our choices, the correct response is "The correct solution set is not included among the other choices."

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Example Question #24 : Pre Calculus

$\textup{Which of the following expressions is equivalent to }\sqrt{\frac{x^{4}y^{11}z^{5}}{x^{2}y^{5}z^{7}}} \textup{ ?}$

Possible Answers:

$\frac{xy^{3}}{z}$

$\frac{xy^{4}}{z}$

$\frac{xy^{3}}{z^{2}}$

$\textup{None of the other answers}$

$\frac{x^{2}y^{3}}{x^{2}}$

Correct answer:

$\frac{xy^{3}}{z}$

Explanation:

$\textup{Reduce the fraction by subtracting exponents, then halve all exponents to find the square root.}$

$\sqrt{\frac{x^{4}y^{11}z^{5}}{x^{2}y^{5}z^{7}}} =\sqrt{\frac{x^{2}y^{6}}{z^{2}}}\:\:=\frac{xy^{3}}{z}$

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High School Math : Functions and Graphs

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #21 : Pre Calculus

Example Question #21 : Pre Calculus

Example Question #1 : Solving Logarithms

Example Question #2 : Solving Logarithms

Example Question #23 : Pre Calculus

Example Question #24 : Pre Calculus

All High School Math Resources

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