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pogonyaev
3 years ago
12

To evaluate log2(3), Autumn reasoned that since log2(2) = 1 and log2(4) = 2, log2(3) must be the average of 1 and 2 and therefor

e log2(3) = 1.5. Use the definition of logarithm to
show that log2(3) cannot be 1.5. Why is her thinking not valid?
Mathematics
1 answer:
geniusboy [140]3 years ago
7 0

Answer:

log₂(3) = 1.585 ≠ 1.5

Her thinking is not valid because the technique of average is valid only  if the graph of the function is a straight line, but the graph of the log function is not a straight line.

Therefore the values cannot be taken by average

Step-by-step explanation:

Given:

log₂(2) = 1

log₂(4) = 2

To evaluate :  log₂(3)

Now,

we know that

logₓ(y) = \frac{\log(y)}{\log(x)}        (Here the log has same base in the numerator and the denominator i.e 10)

therefore,

log₂(3) =  \frac{\log(3)}{\log(2)}

also,

log(2) = 0.3010

log(3) = 0.4771

thus,

log₂(3) =  \frac{0.4771}{0.3010}

or

log₂(3) = 1.585 ≠ 1.5

Her thinking is not valid because the technique of average is valid only  if the graph of the function is a straight line, but the graph of the log function is not a straight line.

Therefore the values cannot be taken by average

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Answer:

5/6

Step-by-step explanation:

When adding fractions, you must ensure the denominator is the same in both fractions.

In this case, the 3 can be multiplied by 2 to equal 6, the other denominator.

When multiplying fractions to create a common denominator, you must multiply the both the numerator and the denominator by the same value, to ensure that the fraction is still equivalent.

2/3 × 2/2 = (2×2)/(3×2) = 4/6

Replace 2/3 with its equivalent 4/6.

Now you will add the numerators together.

1/6 + 4/6 = (1+4)/6 = 5/6

Your final answer is 5/6

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Answer:

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Step-by-step explanation:

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Round this to the nearest inch:

56.6

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\boxed{\underline{\bf \: ANSWER}}

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\dashrightarrow \sf 5\times 5+5\times \left(-6i\right)+6i\times 5+6\left(-6\right)\left(-1\right)

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Now cancel -30i & +30i

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Now the equation becomes

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________

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