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(b) 1

Hint. x = log_ba ⇒ x = \(\frac{log_e\,a}{log_e\,b}\), y = log_c b ⇒ y = \(\frac{log_e\,b}{log_e\,c}\), 

z = log_ac ⇒ z = \(\frac{log_e\,c}{log_e\,a}\).

Given, log₃2, log₃(2^x – 5) and log₃(2^x – \(\frac{7}{2}\)) are in A.P.

⇒ 2[log₃(2^x-5)] = log₃ 2 + log₃ \(\big(2^x-\frac{7}{2}\big)\)

⇒ log₃ (2x – 5)² = log₃ [2 × \(\big(2^x-\frac{7}{2}\big)\)] 

⇒ (2^x – 5)² = (2^{x + 1} – 7) ⇒ 2^2x – 10.2^x + 25 = 2.2^x – 7 

⇒ 2^2x – 12.2^x + 32 = 0 ⇒ y² – 12y + 32 = 0 [Let y = 2^x ] 

⇒ (y – 8) (y – 4) = 0 ⇒ y = 8 or 4 ⇒ 2^x = 8 or 2^x = 4 ⇒ x = 3 or 2

(c) \(\frac{log(x-y)}{log(x+y)}\)

Given, (x⁴ – 2x²y² + y²)^{a –1} = (x – y)^2a (x + y)^–2 

⇒ [(x² – y²)²]^{a –1} = (x – y)^2a (x + y)^–2 

⇒ (x – y)^{2(a – 1)} (x + y)^{2(a –1)} = (x – y)^2a (x + y)^–2

⇒ \(\frac{(x-y)^{2(a-1)}}{(x-y)^{2a}}\) . \(\frac{(x+y)^{2(a-1)}}{(x+y)^{-2}}\) = 1 ⇒ (x-y)^-2 (x+y)^2a = 1

⇒ log [(x – y)^–2 (x + y)^2a] = log 1 ⇒ –2 log (x – y) + 2a log(x + y) = log 1

⇒ 2a log (x + y) = 2 log (x – y) ⇒ a = \(\frac{log(x-y)}{log(x+y)}\).                               [Since log 1 = 0]

(d) \(\frac{1}{α + β+γ}\)

α = \(\frac{1}{log_ax}\), β = \(\frac{1}{log_bx}\), γ = \(\frac{1}{log_cx}\)

⇒ α = log_xa, β = log_xb, γ = log_xc 

⇒ α + β + γ = log_xa + log_xb + log_xc = log_x(abc)

⇒ \(\frac{1}{α + β+γ}\) = \(\frac{1}{log_x(abc)}\) = log_abcx.

(a) 0

Let h be the first term and k be the common ratio of a GP, then 

a = hk^{p – 1}, b = hk^{q – 1}, c = hk^{r – 1}

∴ (q – r) log a + (r – p) log b + (p – q) log c 

= log [hk^{p –1}]^{q – r} + log [hk^{q –1}]^{r – p} + log[hk^{r –1}]^{p – q} 

= log(h^{q – r + r – p + p – q}) (k^{p – 1})^{q – r} (k^{q –1})^{r – p} (k^{r –1})^{p – q} 

= log(h^o k^o) = log 1 = 0.

(a) log_a (log_ba) 

If log_x a, a^x/2 and log_bx are in GP, then (a^x/2)²  = (log_bx) × (log_x a) 

⇒ a^x = log_ba ⇒ log a^x = log (log_ba) ⇒ x log a = log (log_ba) ⇒ x log_a a = log_a (log_ba) 

⇒ x = log_a (log_ba).

According to question,

⇒ 1/log_ax + 1/log_cx = 2/log_bx

⇒ log_xa + log_xc= 2 log_xb

⇒ log_x(ac) = log_x(b)²

On comparing, ac = b² or b² = ac

Hence, a, b, c are in G.P. and G.M. between a and c is b.

(b) log_a (log_e a) – log_a (log_e b) 

 log_xa, \(a^{\frac{x}{2}}\), log_bx are in GP ⇒ \(\big[a^{\frac{x}{2}}\big]^2\) = log_xa . log_bx

⇒ a^x = \(\frac{\text{log}\,a}{\text{log}\,x}.\frac{\text{log}\,x}{\text{log}\,b}\)

⇒ a^x = \(\frac{\text{log}\,a}{\text{log}\,b} = \text{log}_b\,a\) 

⇒ x = log_a (log_ba)

= log a = \(\frac{\text{log}_e\,a}{\text{log}_e\,b}\) = log_a (log_e a) – log_a (log_e b) 

(c) exactly three real solutions 

Given, \(x^{\frac{3}{4}(\text{log}_2\,x)^2}\) \(+\text{log}_2\,x-\frac{5}{4}\) = √2

Taking log to the base 2 of both the sides, we have

\(\bigg[\frac{3}{4}^{(\text{log}_2\,x)^2+(\text{log}_2)-\frac{5}{4}}​​\bigg]\) log₂ x = log₂ √2 = log₂ \(2^{\frac{1}{2}}\) = \(\frac{1}{2}\)log₂ 2 = \(\frac{1}{2}\)

Let us assume log₂x = a. Then,

\(\bigg(\frac{3}{4}a^2+a-\frac{5}{4}\bigg)a\) = \(\frac{1}{2}\) ⇒ 3a³ + 4a² - 5a = 2

⇒ 3a³ + 4a² – 5a – 2 = 0. 

Using hit and trial method check for a = 1. 

f(a) = 3a³ + 4a² – 5a – 2 ⇒ f(1) = 3.1³ + 4.1² – 5.1 – 2 = 0 

∴ (a – 1) is a factor of 3a³ + 4a² – 5a – 2

∴ Now by dividing 3a³ + 4a² – 5a – 2 by (a – 1), we get 

3a³ + 4a² – 5a – 2 = (a – 1) (3a + 1) (a + 2) = 0

⇒ a = 1 or a = \(-\frac{1}{3}\) or a = - 2

⇒ log₂x = 1 or log₂x = \(-\frac{1}{3}\) or log₂x = - 2

⇒ x = 2¹ = 2 or x = 2^-1/3 or x = 2^-2 = \(\frac{1}{4}\)

∴ The given equation has exactly three real solutions, wherein x = 2^–1/3 is irrational.

(d) x

log_am = x ⇒ ax = m 

log_1/a \(\frac{1}{m}\) = y ⇒ (\(\frac{1}{a}\))y = \(\frac{1}{m}\) ⇒ m = ay ⇒ ay = ax ⇒ y = x

(c) 100

5^{log x} – 3^{log x –1} = 3^{log x + 1} – 5^{log x –1} 

⇒ 5^{log x} – 3^{log x} x 3^–1 = 3^{log x} . 3 – 5^{log x} . 5^–1

⇒ 5^{log x} – \(\frac{1}{3}\)3^{log x} = 3 x 3^{log x} - \(\frac{1}{5}\) x 5^{log x}

⇒ \(\bigg(3+\frac{1}{3}\bigg)\)3^{log x} = \(\bigg(5+\frac{1}{5}\bigg)\)5^{log x} ⇒ \(\frac{10}{3}\) x 3^{log x =} \(\frac{6}{5}\)x 5^{log x}

⇒ \(\frac{3^{log\,x}}{5^{log\,x}}\) = \(\frac{6}{5}\) x \(\frac{3}{10}\) = \(\frac{9}{25}\) ⇒ \(\big(\frac{3}{5}\big)^{log\,x}\) = \(\big(\frac{3}{5}\big)^2\)

⇒ log₁₀ x = 2 ⇒ x = 10² = 100

(d) 4

Hint. Given exp. =  log₁₀ \(\bigg\{10^4\bigg(\frac{a+10b+10^2c}{a+10b+10^2c}\bigg)\bigg\}\) 

(c) 2

Given exp. = log 6 + 2 log5 + log 4 – log 3 – log 2 

= log 6 + log (5)² + log 4 – log 3 – log 2 

= log 6 + log 25 + log 4 – (log 3 + log 2) 

= log (6 × 25 × 4) – log (3 × 2)

= log \(\bigg(\frac{6\times25\times4}{3\times2}\bigg)\) = log₁₀ 100 = log₁₀ 10²

= 2 log₁₀ 10 = 2 x 1 = 2.

(d) 0

log\(\frac{a^2}{bc}\) + log\(\frac{b^2}{ac}\) + log\(\frac{c^2}{ab}\) = log\(\bigg(\)\(\frac{a^2}{bc}\) x \(\frac{b^2}{ac}\) x \(\frac{c^2}{ab}\) \(\bigg)\)

= log \(\bigg(\frac{a^2b^2c^2}{a^2b^2c^2}\bigg)\) log 1 = 0.

(d) 1 - m + n.

Given, log_r 6 = m and log_r 3 = n 

Since, log_r 6 = log_r (2 × 3) = log_r 2 + log_r 3 

⇒ log_r 2 + log_r 3 = m 

⇒ log_r 2 + n = m ⇒ log_r 2 = m – n 

Now, logr (\(\frac{r}{2}\)) = log_r r – log_r 2 = 1 – (m – n) = 1 – m + n.