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If a = log12 24, b = log36 24 ,b = log48 36 then prove that 1 + abc =
- a)2bc
- b)2ca
- c)2ba
- d)3bc
Correct answer is option 'A'. Can you explain this answer?
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If a = log12 24, b = log36 24 ,b = log48 36then prove that 1 + abc =a)...
Proof:
Given, a = log12 24, b = log36 24, and c = log48 36
We need to prove that 1/abc = 2bc
Let's start by expressing a, b, and c in terms of a common base, say 2.
a = log12 24
Using the change of base formula, we get:
a = log2 24 / log2 12
a = log2 (2 x 2 x 2 x 3) / log2 (2 x 2 x 3)
a = (log2 2 + log2 2 + log2 2 + log2 3) / (log2 2 + log2 2 + log2 3)
a = (3 + log2 3) / (2 + log2 3)
Similarly,
b = log36 24
Using the change of base formula, we get:
b = log2 24 / log2 36
b = log2 (2 x 2 x 2 x 3) / log2 (2 x 2 x 3 x 3)
b = (log2 2 + log2 2 + log2 2 + log2 3) / (log2 2 + log2 2 + log2 3 + log2 3)
b = (3 + log2 3) / (2 + 2log2 3)
c = log48 36
Using the change of base formula, we get:
c = log2 36 / log2 48
c = log2 (2 x 2 x 3 x 3) / log2 (2 x 2 x 2 x 2 x 3)
c = (log2 2 + log2 2 + log2 3 + log2 3) / (log2 2 + log2 2 + log2 2 + log2 2 + log2 3)
c = (2 + 2log2 3) / (4 + log2 3)
Now, let's use these expressions to simplify 1/abc:
1/abc = 1 / (log12 24 x log36 24 x log48 36)
Substituting the values of a, b, and c, we get:
1/abc = 1 / [(log2 24 / log2 12) x (log2 24 / log2 36) x (log2 36 / log2 48)]
1/abc = 1 / [(log2 2 x log2 2 x log2 2 x log2 3) / (log2 12 x log2 3 x log2 3 x log2 2)]
1/abc = (log2 12 x log2 3 x log2 3 x log2 2) / (log2 2 x log2 2 x log2 2 x log2 3)
1/abc = (log2 12 / log2 2) x (log2 3 / log2 2) x (log2 3 / log2 2) x (log2 2 / log2 3)
1/abc = 2 x 3 x 3/2 x
Given, a = log12 24, b = log36 24, and c = log48 36
We need to prove that 1/abc = 2bc
Let's start by expressing a, b, and c in terms of a common base, say 2.
a = log12 24
Using the change of base formula, we get:
a = log2 24 / log2 12
a = log2 (2 x 2 x 2 x 3) / log2 (2 x 2 x 3)
a = (log2 2 + log2 2 + log2 2 + log2 3) / (log2 2 + log2 2 + log2 3)
a = (3 + log2 3) / (2 + log2 3)
Similarly,
b = log36 24
Using the change of base formula, we get:
b = log2 24 / log2 36
b = log2 (2 x 2 x 2 x 3) / log2 (2 x 2 x 3 x 3)
b = (log2 2 + log2 2 + log2 2 + log2 3) / (log2 2 + log2 2 + log2 3 + log2 3)
b = (3 + log2 3) / (2 + 2log2 3)
c = log48 36
Using the change of base formula, we get:
c = log2 36 / log2 48
c = log2 (2 x 2 x 3 x 3) / log2 (2 x 2 x 2 x 2 x 3)
c = (log2 2 + log2 2 + log2 3 + log2 3) / (log2 2 + log2 2 + log2 2 + log2 2 + log2 3)
c = (2 + 2log2 3) / (4 + log2 3)
Now, let's use these expressions to simplify 1/abc:
1/abc = 1 / (log12 24 x log36 24 x log48 36)
Substituting the values of a, b, and c, we get:
1/abc = 1 / [(log2 24 / log2 12) x (log2 24 / log2 36) x (log2 36 / log2 48)]
1/abc = 1 / [(log2 2 x log2 2 x log2 2 x log2 3) / (log2 12 x log2 3 x log2 3 x log2 2)]
1/abc = (log2 12 x log2 3 x log2 3 x log2 2) / (log2 2 x log2 2 x log2 2 x log2 3)
1/abc = (log2 12 / log2 2) x (log2 3 / log2 2) x (log2 3 / log2 2) x (log2 2 / log2 3)
1/abc = 2 x 3 x 3/2 x
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