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Algebra

Use of the Laws of Logarithms: Set – III

Science > Mathematics > Algebra > Logarithms > Use of Laws of Logarithms Set – III

In the last article, we have studied to solve problems on the use of laws of logarithms to prove given logarithmic expression. In this article, we shall study to solve more problems on these laws to prove given relation using given logarithmic expression.

Example 01:

Use of Laws of Logarithms

Solution:

Given

Use of Laws of Logarithms

Dividing both sides by xy

Use of Laws of Logarithms

Proved as required.

Example 02:

Use of Laws of Logarithms

Solution:

Given

Use of Laws of Logarithms

Dividing both sides by xy

Use of Laws of Logarithms

Proved as required.

Example 03:

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Solution:

Given

Use of Laws of Logarithms

Dividing both sides by xy

Use of Laws of Logarithms

Proved as required.

Example 04:

Use of Laws of Logarithms

Solution:

Given

Use of Laws of Logarithms

Proved as required.

Example 05:

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Solution:

Given

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Proved as required.

Example 06:

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Solution:

Given

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Proved as required.

Example 07:

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Solution:

Given

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Proved as required.

Example 08:

Given 2log2(x + y) = 3 + log2x + log 2y, Show that x2 + y2 = 6xy

Solution:

Given 2log2(x + y) = 3 + log2x + log 2y

log2(x + y)2 = 3 log22 + log2x + log 2y

log2(x + y)2 =  log223 + log2x + log 2y

log2(x + y)2 =  log28 + log2x + log 2y

log2(x + y)2 =  log28xy

 (x + y)2 =  8xy

x2 + 2xy + y2 = 8xy

x2 + y2 = 6xy

Proved as required.

Example 09:

If a2 + b2 = 3ab, show that blank

Solution:

a2 + b2 = 3ab

Adding 2ab on both sides

a2 + 2ab + b2 = 3ab + 2ab

(a + b)2 = 5ab

blank

Proved as required.

Example 10:

If a2 -12ab+ 4b2 = 0, show that

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Solution:

a2 -12ab+ 4b2 = 0

a2 + 4b2 = 12ab

Adding 4ab on both sides

a2 + 4ab + 4b2 = 12ab + 4ab

(a + 2b)2 = 16ab

blank

Proved as required

Example 11:

If a2 + b2 = 7ab, show that blank

Solution:

a2 + b2 = 7ab

Adding 2ab on both sides

a2 + 2ab + b2 = 7ab + 2ab

(a + b)2 = 9ab

blank

Proved as required.

Example 12:

If b2 = ac, prove that log a + log c = 2 log b

Solution:

b2 = ac

Taking log on both sides

Log b2 = log (ac)

2 log b = log a +log c

Log a + log c = 2 log b

Proved as required.

In the next article, we shall study to solve more problems on these laws to prove given logarithmic relation.

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