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valkas [14]
1 year ago
6

If log (m + n) = log m+log n, show that m =n/n-1​

Mathematics
1 answer:
Sedbober [7]1 year ago
3 0

log (m + n) = log m+ log n and proved it m =n/n-1​

Given;

If log (m + n) = log m+ log n

To show that the m =n/n-1​

Now, According to the question:

We know that,

Log (m + n) = log m + log n

Log (m + n ) = log (mn).                 [log a + log b = log ab ]

Cancelling the log on both sides.

then,

m + n = mn

=> n = mn - m

=> n = m (n - 1)

=> m = n / n - 1

Hence Proved

log (m + n) = log m+ log n and proved it m =n/n-1​

What is Logarithm?

A logarithm is the power to which a number must be raised in order to get some other number (see Section 3 of this Math Review for more about exponents). For example, the base ten logarithm of 100 is 2, because ten raised to the power of two is 100: log 100 = 2.

Learn more about Logarithm at:

brainly.com/question/16845433

#SPJ1

 

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Minus each number to figure out the common difference which is -4.

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3 years ago
Order the side lengths from least to greatest
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Answer:

BC < CE < BE < ED < BD

Step-by-step explanation:

In the triangle BCE,

m∠BEC + m∠BCE + m∠CBE = 180°

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m∠BEC = 180 -  135

m∠BEC = 45°

Order of the angles from least to greatest,

m∠BEC < m∠CBE > mBCE

Sides opposite to these sides will be in the same ratio,

BC < CE < BE ----------(1)

Now in ΔBED,

m∠BEC + m∠BED = 180°

m∠BED = 180 - 45

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Now, m∠BDE + m∠BED + DBE = 180°

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m∠DBE = 180 - 146

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Order of the angles from least to greatest will be,

∠BDE < ∠DBE < ∠BED

Sides opposite to these angles will be in the same order.

BE < ED < BD ----------(2)

From relation (1) and (2),

BC < CE < BE < ED < BD

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