Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.12 Text Book Back Questions and Answers, Notes.

## 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.

Answer:

Given y = b^{x} ⇒ log_{b}y = x, x ∈ R with range (0, ∞) (-∞, ∞)

Question 2.

Compute log_{9} 27 – log_{27} 9

Answer:

log_{9}27 – log_{27}9 = log_{9} 3^{3} – log_{27} 3^{2}

= 3 log_{9} 3 – 2 log_{27} 3 —— (1)

By change of base rule [log_{b} a = \(\frac{1}{\log _{a} b}\)]

Question 3.

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

Answer:

Question 4.

Solve log _{4 }2 ^{8x} = 2 ^{log28}

Answer:

Given log _{4 }2^{8x} = 2 ^{log28}

log _{4 }2^{8x} = 2^{log223}

log _{4 }2^{8x} = 2^{3 log22}

log _{4 }2^{8x} = 2^{3} = 8

2^{8x} = 4^{8}

(2^{2})^{4x} = 4^{8}

⇒ (4)^{4x} = 4^{8}

⇒ 4x = 8

⇒ x = \(\frac { 8 }{ 4 }\) = 2

Question 5.

If a^{2} + b^{2} = 7ab, show that

log \(\left(\frac{a+b}{3}\right)\) = \(\frac{1}{2}\) (log a + log b)

Answer:

Given

a^{2} + b^{2} = 7ab

Adding both sides 2ab we get

a^{2} + b^{2} + 2ab = 7ab + 2ab

(a + b)^{2} = 9ab

Taking square root on both sides

Taking logarithm on both sides

Question 6.

Prove that log \(\frac{\mathbf{a}^{2}}{\mathbf{b c}}\) + log \(\frac{\mathbf{b}^{2}}{\mathbf{c a}}\) + log \(\frac{c^{2}}{a b}\) = 0

Answer:

Question 7.

Prove that

Answer:

Question 8.

Prove that log_{a2} a + log_{b2} b + log_{c2} c = \(\frac{1}{8}\)

Answer:

Question 9.

Prove log a + log a^{2} + log a^{3} + ……… + log a^{n} = \(\frac{n(n+1)}{2}\) log a

Answer:

log a + log a^{2} + log a^{3} + ……… + log a^{n}

= log a + 2 log a + 3 log a + ………….. + n log a

= log a (1 + 2 + 3 + ………….. + n)

= log a × \(\frac{n(n+1)}{2}\)

= \(\frac{n(n+1)}{2}\) log a

Question 10.

If , then prove that xyz = 1

Answer:

Let \(\frac{\log x}{y-z}\) = k

log x = k(y – z)

log x = ky – kz ——— (1)

Similarly log y = k(z – x) = kz – kx ——(2)

log z = k(x – y) = kx – ky ——- (3)

Adding (1), (2) and (3)

log x + log y + log z = ky – kz + kz – kx + kx – ky

log (xyz) = 0 = log 1

xyz = 1

Question 11.

Solve log_{2}x – 3 log_{1/2}x = 6

Answer:

Question 12.

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

Answer:

log_{5 – x}(x^{2} – 6x + 65) = 2

⇒ x^{2} – 6x + 65 = (5 – x)^{2}

⇒ x^{2} – 6x + 65 = 25 + x^{2} – 10x

⇒ x^{2} – 6x + 65 – 25 – x^{2} + 10x = 0

⇒ 4x + 40 = 0

⇒ 4x = -40

⇒ x = -10