Logaritmo

por **karenblond** » Qui Mai 26, 2011 13:34

[url][URL=http://imageshack.us/photo/my-images/855/logx.png/] Imagem

[/url]
[/url]

por **Molina** » Qui Mai 26, 2011 14:15

Boa tarde, Karen.

Em 2013 teremos t = 2013:

$f(t)=5+log_2\left( \frac{t - 1997}{8} \right)$

$f(2013)=5+log_2\left( \frac{2013 - 1997}{8} \right)$

$f(2013)=5+log_2\left( \frac{16}{8} \right)$

$f(2013)=5+log_2\left( 2 \right)$

f(2013)=5+1=6

E em 2005 tivemos t = 2005:

$f(t)=5+log_2\left( \frac{t - 1997}{8} \right)$

$f(2005)=5+log_2\left( \frac{2005 - 1997}{8} \right)$

$f(2005)=5+log_2\left( \frac{8}{8} \right)$

$f(2005)=5+log_2\left( 1 \right)$

f(2005)=5+0=5