Question

2 3 n
S= log x + log X + log X+…+log X
3 3 3 3
X>0, n€N

Answer 1

Răspuns:

Explicație pas cu pas:

Metaoda 1: folosim proprietatea logaritmului:

$log_{a} x^b = b*log_{a} x$

asadar:

$log_{3} x + log_{3} x^2 + log_{3} x^3 + ... + log_{3} x^n =$

$=log_{3} x + 2*log_{3} x + 3*log_{3} x + ... + n*log_{3} x =$

$=log_{3} x *(1 + 2 + 3 + ... + n)=$

$=log_{3} x *\frac{n*(n+1)}{2}= \frac{n*(n+1)}{2}*log_{3} x = log_{3} x ^\frac{n*(n+1)}{2}$

Metaoda 2: folosim proprietatea logaritmului:

$log_{a} x^b = b*log_{a} x$

asadar:

$log_{3} x + log_{3} x^2 + log_{3} x^3 + ... + log_{3} x^n =$

$=log_{3} x + 2*log_{3} x + 3*log_{3} x + ... + n*log_{3} x =$

$=log_{3} x *(1 + 2 + 3 + ... + n)=$

$= log_{3} x ^\frac{n*(n+1)}{2} = \frac{n*(n+1)}{2}*log_{3} x$

Answer 2

Răspuns:

Explicație pas cu pas: