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

Answer: (d)

Semiperimeter of triangle = ${30 + 16 + 28}/2$ = 37

P – a = 7

P – b = 9

P – c = 21

P = 37

∴ quantities that are required

= 37, 21, 9, 7

Answer: (a)

$log_{10}(3/2) + log_{10}(4/3) + log_{10}(5/4)+ ..... + 8^{th} term$

= $log_{10}(3/2) + log_{10}(4/3)+ log_{10}(5/4)+ ..... +log_{10}({10}/9)$

= $log_{10}(3/2 × 4/3 × 5/4 ×......× {10}/9)$

= $log_{10}({10}/2)log_{10} 5$.

Answer: (c)

${\text"log"_8}x +{ \text"log"_8} {1/6} = 1/3$

or ${\text"log"_8} (x × {1/6}) = 1/3 or {\text"log"_8} (x/6) = 1/3$

or $x/6 = (8)^{1/3}$ {∵ $\text"log"_a$ b = x⇔(a)x=b}

or ${x/6} = (2^3)^{1/3}$ or x = 12

Answer: (a)

$log (108)^{10} = 10 log 108 = 10 log (2^2 × 3^3)$ = 10 (2log2 + 3log3)

= 10 (2 × 0.301 + 3 × 0.477 ) = 10 (.602 + 1.431)

= 10 × 2.033 = 20.33

integral part = 20

No. of digits = 20 + 1 = 21

Answer: (c)

$log_{10} [1 - [1 - (1 - x^2)^{-1}]^{-1}]^{-1/2}$ = 1

⇒$log_{10}[1 - [1 - 1/{1 - x^2}]^{-1}]^{-1/2}$ = 1

⇒$log_{10}[1 - [{- x^2}/{1 - x^2}]^{-1}]^{- 1/2}$ = 1

⇒$log_{10}[1 - {(1 - x^2)}/{- x^2}]^{-1/2}$ = 1

⇒$log_{10}[{-x^2 - (1 - x^2)}/{- x^2}]^{-1/2}$ = 1

⇒$log_{10}[1/{x^2}]^{-1/2}$ = 1

⇒$log_{10} x = log_{10}10$

x = 10

⇒$log_{10}$