doubt in logarithm

Question

can you please explain how 1-(1/2)^ln(n)  becomes (n-1)/n ? 

Answer will be $\Theta(n)$
\[
\begin{aligned}
j & =n / 2+n / 4+n / 8+\ldots+1 \\
& =n\left[1 / 2^{1}+1 / 2^{2}+1 / 2^{3}+\ldots+1 / 2^{\lg n}\right]
\end{aligned}
\]
(Sum of first $n$ terms of GP is $\left[a \cdot \frac{1-r^{r}}{1-r}\right]$, where $a$ is the first term, $r$ is the common ratio $<1$, and $n$ is the number of terms)<br /> \[
\begin{array}{l}
=n\left[1 / 2 \frac{1-(1 / 2)^{18 n}}{1-1 / 2}\right] \\
=n\left[\frac{n-1}{n}\right] \Leftarrow \\
=n-1=\Theta(n)
\end{array}
\]

Comments

1-$({\frac{1}{2}})^{lg n}$ = 1-$({2^{-1}})^{lg n}$ = 1-$({2})^{-lg n}$ = 1-$({2})^{{lg\; (n}^{-1})}$ =1-$({n^{-1}})^{lg 2}$ =  1-$({n^{-1}})^{1}$ = 1-$({n^{-1}})$ = 1-$\frac{1}{n}$ = $\frac{n-1}{n}$

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see (1/2)^log₂n 

can be writen as... 1/ 2^log₂n

= 1/n^log₂2 = (1/n)   (n^log_ab <==> b^log_an) this is the formulae...

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Thank You @Shaik Masthan  , You have answered many of my doubts, Your help is truly invaluable.

Regard

Answer 1

could you post full question, just curious

Comments

https://gateoverflow.in/976/gate2006-15

Answer 2

Here 1/2 is cancel out by 1 - 1/2

$n[1 - (1/2) ^\ log n]$

log property ::   2 ^ log n  (base 2)

                         then replace n with 2

                      = n ^ log 2 = n

So, our equestion are in form ->n[1 - ( 2^(-1) ) ^logn]

 

                        so we can write as n[1 - (n^-1)  ^ log2 ]                                                  then

                         =    $n[ 1- 1/n ]$

                        =   $n[ \frac{n-1}{n} ]$

                         = $ n - 1 $