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High School Math : Simplifying Logarithms

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Example Question #1 : Properties Of Logarithms

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

Possible Answers:

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

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

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

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

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

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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High School Math : Simplifying Logarithms

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Example Question #1 : Properties Of Logarithms

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