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4 years ago

Find f(2), f(0), f(-1)

Mathematics

1 answer:

mr Goodwill [35]4 years ago

7 0

Put the values of x to the equation of the function f:

$f(x)=\dfrac{x+21}{9x-3}\\\\f(2)=\dfrac{2+21}{9(2)-3}=\dfrac{23}{18-3}=\dfrac{23}{15}\\\\f(0)=\dfrac{0+21}{9(0)-3}=\dfrac{21}{-3}=-7\\\\f(-1)=\dfrac{-1+21}{9(-1)-3}=\dfrac{20}{-9-3}=\dfrac{20}{-12}=-\dfrac{5}{3}$

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write the logarithm as a sum or difference of logarithms. simplify each term as much as possible. log4(6ab)

Dominik [7]

The logarithm written as a sum of logarithm and simplified as much as possible is $\frac{1}{2} +log_{4 }3 + log_{4 }a + log_{4 }b$

<h3>Simplifying Logarithms</h3>

From the question, we are to write the given logarithm expression as a sum or difference of logarithms

The given logarithm is

$log_{4 }6ab$

This can be written as

$log_{4 }6 \times a \times b$

From one of the rules of logarithm, we have that

$log_{x }yz= log_{x }y + log_{x }z$

Thus,

$log_{4 }6 \times a \times b$ becomes

$log_{4 }6 + log_{4 }a + log_{4 }b$

This can be further simplified into

$log_{4 }3 \times 2 + log_{4 }a + log_{4 }b$

$log_{4 }3 + log_{4 }2 + log_{4 }a + log_{4 }b$

If desired, this can be further simplified into

$log_{4 }3 + log_{2^{2} }2 + log_{4 }a + log_{4 }b$

$log_{4 }3 + \frac{1}{2} log_{2}2 + log_{4 }a + log_{4 }b$

$log_{4 }3 + \frac{1}{2} (1)+ log_{4 }a + log_{4 }b$

$\frac{1}{2} +log_{4 }3 + log_{4 }a + log_{4 }b$

Hence, the logarithm written as a sum of logarithm and simplified as much as possible is $\frac{1}{2} +log_{4 }3 + log_{4 }a + log_{4 }b$

Learn more on Simplifying logarithms here: brainly.com/question/17851187

#SPJ1

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