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

#### LOGARITHM PRACTICE TEST [1 - EXERCISES]

Question : 1

What is the value of $(\text"log"_{1/2} 2) (\text"log"_{1/3}3) (\text"log"_{1/4}4).....(\text"log"_{1/{1000}}1000)$

a) 1 or – 1

b) 1

c) 0

d) – 1

e) None of these

$(\text"log"_{1/2}2)(\text"log"_{1/3}3)(\text"log"_{1/4}4).......(\text"log"_{1/{1000}}1000)$

=$({\text"log" 2}/{\text"log"{1/2}})({\text"log" 3}/{\text"log"{1/3}})({\text"log" 4}/{\text"log"{1/4}})......({\text"log" 1000}/{\text"log"{1/{1000}}})$ $(∵ \text"log"_b a = {\text"log" a}/{\text"log" b})$

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

= (-1) × (-1) × (-1) × .....× (-1)

(∵ number of terms is odd) = -1

Question : 2

If log x = 1.2500 and y = $x^{logx}$ , then what is log y equal to?

a) 1.5625

b) 4.2500

c) 2.5625

d) 1.2500

We have, y = $x^{logx}$

Taking log on both sides log

y = log $(x^{logx})$

= logx. logx = $(1.25)^2$ = 1.5625

Question : 3

If a = $\text"log"_{24} 12, b = \text"log"_{36}$ 24, C = $\text"log"_{48}$ 36. Then 1 + abc is equal to

a) 2ab

b) 2ac

c) 2abc

d) 2bc

e) None of these

abc = ${\text"log" 12}/{\text"log" 24} . {\text"log" 24}/{\text"log" 36} . {\text"log" 36}/{\text"log" 48} = {\text"log" 12}/{\text"log" 48}$

∴ 1 + abc = ${\text"log" 48 + \text"log"12}/{\text"log" 48} = {\text"log" (48.12)}/{\text"log" 48}$

= ${\text"log"24}^2/{\text"log" 48} = 2.{\text"log" 24}/{\text"log" 48}$ = 2bc

Question : 4

If ${\text"log"_a} b = 1/2, {\text"log"_b}$ c = $1/3$ and ${\text"log"_c}$ a = $k/5$, then the value of k is

a) 30

b) 25

c) 20

d) 35

e) None of these

${\text"log"_a} b = {1/2}, {\text"log"_b}c= {1/3}$, and ${\text"log"_c} a = k/5$

⇒${\text"log" b}/{\text"log" a} = {1/2}, {\text"log" c}/{\text"log" b} = {1/3}, {\text"log" a}/{\text"log" c} = k/5$

⇒${1/2} × {1/3} × {k/5}$ = 1⇒k = 30

Question : 5

What is the solution of the equation x $log_{10} ({10}/3) + log_{10} 3 = log_{10} (2 + 3^x)$ + x ?

a) 1

b) 10

c) 3

d) 0

$x log_{10}({10}/3) + log_{10} 3 = log_{10} (2 + 3^x) + x$

$xlog_{10} 10 - x log_{10} 3 + log_{10} 3 = log_{10} (2 + 3^x)$ + x

x - $log_{10^3^x} + log_{10^3} = log_{10} (2 + 3^x)$ + x

$log_{10[3/{3^x}]} log_{10} (2 + 3^x)$

⇒$3^{1–x} = 2 + 3^x$

⇒$3^{1–x} –3^x = 3^1 – 3^0$

x = 0

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