Exponential and Logarithmic Functions - High School Math

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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 .

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