# $$\frac{1}{{{{\log }_2}x}} + \frac{1}{{{{\log }_3}x}} + \frac{1}{{{{\log }_4}x}} + \ldots .. + \frac{1}{{{{\log }_{50}}x}},x \ne 1$$ is equal to

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1. $$\frac{{50}}{{{{\log }_{50}}x}}$$
2. $$\frac{{49}}{{{{\log }_{49}}x}}$$
3. $$\frac{{1}}{{{{\log }_{50!}}x}}$$
4. $$\frac{{1}}{{{{\log }_{49!}}x}}$$

Option 3 : $$\frac{{1}}{{{{\log }_{50!}}x}}$$
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CT 1: हिन्दी (आदिकाल)
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10 Questions 40 Marks 10 Mins

## Detailed Solution

Concept:

We know that,

$${\log _x}y = \frac{1}{{{{\log }_y}x}}$$ ---(1)

Calculation:

Given:

$$\frac{1}{{{{\log }_2}x}} + \frac{1}{{{{\log }_3}x}} + \frac{1}{{{{\log }_4}x}} + \ldots .. + \frac{1}{{{{\log }_{50}}x}},x \ne 1$$

By using above property, It can be written as:

$${\log _x}2 + {\log _x}3 + \ldots + {\log _x}50\;\;\;\;\;\;\;\;\;\;$$

$${\log _x}\left[ {2 \times 3 \times 4 \times 5 \times \ldots \times 50} \right]$$

$${\log _x}50! \Rightarrow \frac{1}{{{{\log }_{50!}}x}}$$

Important Points

Properties of Logarithms:

1. $${\log _a}a = 1$$
2. $${\log _a}\left( {x.y} \right) = {\log _a}x + {\log _a}y$$
3. $${\log _a}\left( {\frac{x}{y}} \right) = {\log _a}x - {\log _a}y$$
4. $${\log _a}\left( {\frac{1}{x}} \right) = - {\log _a}x$$
5. $${\rm{lo}}{{\rm{g}}_a}{x^p} = p{\rm{lo}}{{\rm{g}}_a}x$$
6. $$lo{g_a}\left( x \right) = \frac{{lo{g_b}\left( x \right)}}{{lo{g_b}\left( a \right)}}$$