Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.12

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.12

Question 1.
Let b > 0 and b ≠ 1. Express y = bx in logarithmic form. Also, state the domain and range of the logarithmic function.
Solution:
Given y = bx
By the definition of logarithm logb y = x
The domain of a logarithmic function is the set of positive real numbers and the range is the set of real numbers.

Question 2.
Compute log927 – log279.
Solution:
Let log927 = x ⇒ 27 = 9x ⇒ 33 = (32)x = 32x
⇒ 2x = 3 ⇒ x = 3/2
Let log279 = x
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.12 1

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.12

Question 3.
Solve log8x + log4x + log2x = 11
Solution:
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Question 4.
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Solution:
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Question 5.
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Solution:
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Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.12

Question 6.
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Solution:
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Another method
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Question 7.
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Solution:
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Question 8.
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Solution:
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Question 9.
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Solution:
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Question 10.
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Solution:
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Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.12

Question 11.
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Solution:
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Question 12.
Solve log5 – x(x2 – 6x + 65) = 2
Solution:
Given log5 – x (x2 – 6x + 65) = 2
By the definition of logarithm
x2 – 6x + 65 = (5 – x)2
x2 – 6x + 65 = 25 – 10x + x2
10x – 6x + 65 – 25 = 0
4x + 40 = 0
4x = – 40
x = \(-\frac{40}{4}\) = -10
x = -10

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.12 Additional Questions

Question 1.
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Solution:
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Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.12

Question 2.
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Solution:
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Question 3.
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Solution:
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Question 4.
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Solution:
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