Logarithm Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (c)

$(log_{1/2}2)(log_{1/3}3)(log_{1/4}4).......(log_{1/{1000}}1000)$

= $({log 2}/{log_{1/2}})({log 3}/{log_{1/3}})({log 4}/{log_{1/4}})......({log 1000}/{log_(1/{1000})})$

$(∵ log_b a = {log a}/{log b})$

= $({log 2}/{- log 2})({log 3}/{- log 3})({log 4}/{- log 4}).........({log 1000}/{- log 1000})$

= (-1) × (-1) × (-1) × ..... × (-1)

(∵ number of factors is odd)

= –1

Answer: (a)

${\text"log"_a}$(ab) = x

⇒${\text"log" ab}/{\text"log" a} = x⇒{\text"log" b}/{\text"log" a} = x - 1⇒{\text"log" a}/{\text"log" b} = 1/{x - 1}$

Now, ${\text"log"_b}(ab) = {\text"log" ab}/{\text"log" b} = {\text"log" a + \text"log" b}/{\text"log" b}$

= ${\text"log" a}/{\text"log" b} + 1 = {1/{x - 1}} + 1 = {1 + x - 1}/{x - 1} = {x}/{x - 1}$

Answer: (d)

Given, $log_r 6 = \text"m and" log_r$ 3 = n

∵ $log_r 6 = log_r$ (2 × 3)

= $log_r 2 + log_r$ 3

∴ $log_r 3 + log_r$ 2 = m

⇒n + $log_r2$ = m

⇒$log_r$ 2 = m – n

∴ $log_r (r/2) = log_r r - log_r 2$

= 1 – m + n

Answer: (b)

$log_{10} 1050 = log_{10}$(3 × 10 × 35)

= $log_{10} 3 + 1 + log_{10} 35$

⇒ 3.0212 = $log_{10} 3 + 1 + 1.5441$

⇒$log_{10} 3$ = 0.4771

Now, $log_{10} 35 = log {7 × 10}/2$

= $log_{10} 7 + log_{10} - log_{10} 2$

⇒ $log_{10}7$ = 0.8451

Now, A = ${√3}/4 × (7)^2$

$log{10^A^4} = 4 log_{10}A$

= 4 log ${√3 × 7^2}/4$

= $4[1/2 log_3 + 2 log_{10} 7 - 2 log 2]$ = 5.3070

Answer: (a)

Let $log_{10}$ 0.0001 = a

a = $log_{10} {1/{(10)^4}$

= $log_{10} 1 – log_{10} (10)^4$ = 0 – 4 = –4