Question

Solve the above :-

Note :- Quality Answers will be appreciated and wrong will ne deleted ✓
And n = 100! ​

Answer 1

$\large\underline{\sf{Solution-}}$

We know that

$\boxed{ \sf{ \: log_{x}(y) = \dfrac{1}{ log_{y}(x) } }}$

Given that

$\rm :\longmapsto\:\dfrac{1}{ log_{2}(n) } + \dfrac{1}{ log_{3}(n) } + \dfrac{1}{ log_{4}(n) } + - - - + \dfrac{1}{ log_{100}(n) }$

$\rm \: = \: log_{n}(2) + log_{n}(3) + log_{n}(4) + - - - - + log_{100}(n)$

$\rm \: = \: log_{n}(2 \times 3 \times 4 \times - - - - \times 100)$

$\: \: \: \: \: \: \boxed{ \sf{ \because \: log_{a}(x) + log_{a}(y) = log_{a}(xy) }}$

can be rewritten as

$\rm \: = \: \: \: log_{n}(1 \times 2 \times 3 \times \times - - - - \times 100)$

$\rm \: = \: \: \: log_{n}(100!)$

$\rm \: = \: \: \: log_{(100!)}(100!) \: \: \: \: \{ \because \: n \: = \: 100! \: \}$

$\rm \: = \: \: \:1$

$\boxed{ \sf{ \because \: log_{x}(x) = 1 }}$

Hence,

★ If n = 100! then

$\rm :\longmapsto\:\dfrac{1}{log_{2}(n)}+\dfrac{1}{log_{3}(n)}+\dfrac{1}{ log_{4}(n) }+--+\dfrac{1}{log_{100}(n)} =1$

Additional information :-

$\boxed{ \sf{ \:logxy = logx \: + \: logy }}$

$\boxed{ \sf{ \:log \: \frac{x}{y} = logx - logy}}$

$\boxed{ \sf{ \: log {x}^{y} = y \: logx}}$

$\boxed{ \sf{ \: log_{ {x}^{p} }( {x}^{q} ) = \frac{q}{p} }}$

$\boxed{ \sf{ \: {e}^{logx} = x}}$

$\boxed{ \sf{ \: {a}^{ log_{a}(x) } = x}}$