Exponential and Logarithmic Functions - Math

Card 0 of 20

Question

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

Answer

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

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

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Question

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

Answer

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

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

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

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

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

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

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

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

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

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 . This is true, so is a solution.

:

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

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

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

However, negative numbers do not have logarithms, so this equation is meaningless. is not a solution, and 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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Question

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

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

Answer

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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Question

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

Answer

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

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

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Question

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

Answer

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 .

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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

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

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

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

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

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

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

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

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

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 . This is true, so is a solution.

:

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

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

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

However, negative numbers do not have logarithms, so this equation is meaningless. is not a solution, and 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."

Compare your answer with the correct one above

Question

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

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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 .

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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

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

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

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

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

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

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

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

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

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 . This is true, so is a solution.

:

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

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

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

However, negative numbers do not have logarithms, so this equation is meaningless. is not a solution, and 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."

Compare your answer with the correct one above

Question

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

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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 .

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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

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

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

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

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

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

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

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

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

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 . This is true, so is a solution.

:

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

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

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

However, negative numbers do not have logarithms, so this equation is meaningless. is not a solution, and 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."

Compare your answer with the correct one above

Question

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

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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 .

Compare your answer with the correct one above

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