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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 = b^{x} in logarithmic form. Also, state the domain and range of the logarithmic function.

Solution:

Given y = b^{x}

By the definition of logarithm log_{b} 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 log_{9}27 – log_{27}9.

Solution:

Let log_{9}27 = x ⇒ 27 = 9^{x} ⇒ 3^{3} = (3^{2})^{x} = 3^{2x}

⇒ 2x = 3 ⇒ x = 3/2

Let log_{27}9 = x

Question 3.

Solve log_{8}x + log_{4}x + log_{2}x = 11

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Another method

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solve log_{5 – x}(x^{2} – 6x + 65) = 2

Solution:

Given log_{5 – x} (x^{2} – 6x + 65) = 2

By the definition of logarithm

x^{2} – 6x + 65 = (5 – x)^{2}

x^{2} – 6x + 65 = 25 – 10x + x^{2}

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.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution: