Question

Answer it as fast as possible

Answer 1

3.

$\sum\limits_{n=2}^{127} log_{2}((1 + \frac{1}{n} )$

$= \sum\limits_{n=2}^{127} log_{2} \frac{(n + 1)}{n}$

$= \sum\limits_{n = 2}^{127} log_{2}(n + 1) \: \: - \: \: \sum\limits_{n = 2}^{127} log_{2}n$

$= ( log_{2}(128) + log_{2}(127) + ... + log_{2}(3) ) - ( log_{2}(127) + log_{2}(126) + ...... + log_{2}(3) log_{2}(2) )$

$= log_{2}(128)$

4.

${(logx)}^{2} + {(logy)}^{2} - log(xy) \times log( \frac{x}{y} )$

$= {(logx)}^{2} + {(logy)}^{2} -( log(xy) \times log( \frac{x}{y} ))$

$= {(logx)}^{2} + {(logy)}^{2} -( logx + logy) \times \: ( logx - logy)$

$= {(logx)}^{2} + {(logy)}^{2} - {(logx)}^{2} + {(logy)}^{2}$

$= 2 \: {(logy)}^{2}$

Answer 2

n=2

∑

127

log

2

((1+

n

1

)

= \sum\limits_{n=2}^{127} log_{2} \frac{(n + 1)}{n}=

n=2

∑

127

log

2

n

(n+1)

= \sum\limits_{n = 2}^{127} log_{2}(n + 1) \: \: - \: \: \sum\limits_{n = 2}^{127} log_{2}n=

n=2

∑

127

log

2

(n+1)−

n=2

∑

127

log

2

n

= ( log_{2}(128) + log_{2}(127) + ... + log_{2}(3) ) - ( log_{2}(127) + log_{2}(126) + ...... + log_{2}(3) log_{2}(2) )=(log

2

(128)+log

2

(127)+...+log

2

(3))−(log

2

(127)+log

2

(126)+......+log

2

(3)log

2

(2))

= log_{2}(128)=log

2

(128)

4.

{(logx)}^{2} + {(logy)}^{2} - log(xy) \times log( \frac{x}{y} )(logx)

2

+(logy)

2

−log(xy)×log(

y

x

)

= {(logx)}^{2} + {(logy)}^{2} -( log(xy) \times log( \frac{x}{y} ))=(logx)

2

+(logy)

2

−(log(xy)×log(

y

x

))

= {(logx)}^{2} + {(logy)}^{2} -( logx + logy) \times \: ( logx - logy)=(logx)

2

+(logy)

2

−(logx+logy)×(logx−logy)

= {(logx)}^{2} + {(logy)}^{2} - {(logx)}^{2} + {(logy)}^{2}=(logx)

2

+(logy)

2

−(logx)

2

+(logy)

2

= 2 \: {(logy)}^{2}=2(logy)

2