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3 years ago

Equation for (0,6) and (4,5)

Mathematics

1 answer:

kow [346]3 years ago

6 0

Answer:

y=-1/4x+6

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(5-6)/(4-0)

m=-1/4

y-y1=m(x-x1)

y-6=-1/4(x-0)

y-6=-1/4(x)

y-6=-1/4x

y=-1/4x+6

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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

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