The Value of logtan10+logtan20+⋯⋯ | Logarithms Questions & Answers

Q:

It is given that log10 2= 0.301 and log10 3 = 0.477. How many digits are there in ${(108)}^{10}$ ?

A) 19	B) 20
C) 21	D) 22

Answer & Explanation Answer: C) 21

Explanation:

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Q:

If 5x - 17 = -x + 7, then x = ?

A) 4	B) 5
C) 7	D) 9

Answer & Explanation Answer: A) 4

Explanation:

Given equation 5x-17 = -x+7

Add 1x to each side of the equation

5x-17+x = -x+7+x

6x=24

x=4

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Q:

Solve the equation ${(\frac{1}{2})}^{2 x + 1} = 1$ ?

A) -1/2	B) 1/2
C) 1	D) -1

Answer & Explanation Answer: A) -1/2

Explanation:

Rewrite equation as ${(\frac{1}{2})}^{2 x + 1} = {(\frac{1}{2})}^{0}$

Leads to 2x + 1 = 0

Solve for x : x = -1/2

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Q:

If $\log_{7} (2)$ = m, then $\log_{49} (28)$ is equal to ?

A) 1/(1+2m)	B) (1+2m)/2
C) 2m/(2m+1)	D) (2m+1)/2m

Answer & Explanation Answer: B) (1+2m)/2

Explanation:

$\log_{49} (28) = \frac{1}{2 \log_{7} (7 \times 4)}$

= $\frac{1}{2} + \frac{1}{2} (2 \log_{7} (2))$
= $\frac{1}{2} + \log_{7} (2)$
= $\frac{1}{2} + m$
= $\frac{1 + 2 m}{2}$ .

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Q:

If $a^{2} + b^{2} = c^{2}$ , then $\frac{1}{l o g_{c + a} (b)} + \frac{1}{l o g_{c - a} (b)} = ?$

A) 1	B) 2
C) 4	D) 8

Answer & Explanation Answer: B) 2

Explanation:

Given $a^{2} + b^{2} = c^{2}$

Now $\frac{1}{\log_{c + a} (b)} + \frac{1}{\log_{c - a} (b)}$

= $\log_{b} (c + a) + \log_{b} (c - a)$

= $\log_{b} (c^{2} - a^{2})$

= $2 \log_{b} (b) = 2$

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Q:

If log 64 = 1.8061, then the value of log 16 will be (approx)?

A) 1.9048	B) 1.2040
C) 0.9840	D) 1.4521

Answer & Explanation Answer: B) 1.2040

Explanation:

Given that, log 64 = 1.8061

i.e $\log (4^{3}) = 1.8061$

--> 3 log 4 = 1.8061

--> log 4 = 0.6020

--> 2 log 4 = 1.2040

$\Rightarrow \log (4^{2}) = 1.2040$

Therefore, log 16 = 1.2040

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Q:

A fast moving superfast express crosses another pasenger train in 20 seconds. The speed of faster train is 72 km/hr and speeds of slower train is 27 km/h. Also the length of faster ntrain is 100m, then find the length of the slower train if they are moving in the same direction.

A) 100 m	B) 125 m
C) 150 m	D) 175 m

Answer & Explanation Answer: C) 150 m

Explanation:

$T i m e = \frac{S u m o f l e n g t h o f t h e t w o t r a i n s}{D i f f e r e n c e i n s p e e d s}$

$20 = \frac{(100 + x)}{\frac{25}{2}}$

$\Rightarrow$ X=150m

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Q:

For $x \in N, x > 1$ , and $p = \log_{x} (x + 1)$ , $q = \log_{x + 1} (x + 2)$ then which one of the following is correct?

A) p < q	B) p = q
C) p > q	D) can't be determined

Answer & Explanation Answer: C) p > q

Explanation:

$\frac{k}{l} > \frac{k + 1}{l + 1}$ for (k,l) > 0 and k > l

Let k = x+1 and l = x

Therefore, $\frac{x + 1}{x} > \frac{(x + 1) + 1}{(x) + 1}$

(x + 1) > x

Therefore, $\frac{\log (x + 1)}{\log (x)} > \frac{\log (x + 2)}{\log (x + 1)}$

$\Rightarrow \log_{x} (x + 1) > \log_{x + 1} (x + 2)$

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The Value of $[\log (\tan (1^{0})) + \log (\tan (2^{0})) + \dots \dots + \log (\tan (89^{0}))]$ is

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The Value of logtan10+logtan20+⋯⋯+logtan890 is

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The Value of $[\log (\tan (1^{0})) + \log (\tan (2^{0})) + \dots \dots + \log (\tan (89^{0}))]$ is