Question

Sabendo que log 2 = x, log 3 = y e log 5 = z, calcule os logaritmos em função de x, y e z:
a) log 27
b) log 20
c) log 7,5
d) log3^2

Answer 1

Propriedades usadas:

logc a^b = blogc a

logc (a . b) = logc a + logc b

a)

log 27

log 3³

3 log 3

3y

b)

log 20

log 2² . 5

2log 2 + log 5

2x + z

c)

log 7,5

log 3 . 5² . 10^(-1)

- 1 + log 3 + 2log 5

- 1 + y + 2z

d)

log 3²

2log 3

2y

Resposta:

a) 3y

b) 2x + z

c) - 1 + y + 2z

d) 2y

Answer 2

Explicação passo-a-passo:

Lembre-se que:

• $\sf log_{b}~a^n=n\cdot log_{b}~a$

• $\sf log_{b}~(a\cdot c)=log_{b}~a+log_{b}~c$

• $\sf log_{b}~\Big(\dfrac{a}{c}\Big)=log_{b}~a-log_{b}~c$

a)

$\sf log~27=log~3^3$

$\sf log~27=3\cdot log~3$

$\sf \red{log~27=3y}$

b)

$\sf log~20=log~(2^2\cdot5)$

$\sf log~20=log~2^2+log~5$

$\sf log~20=2\cdot log~2+log~5$

$\sf \red{log~20=2x+z}$

c)

$\sf log~7,5=log~\Big(\dfrac{75}{10}\Big)$

$\sf log~7,5=log~\Big(\dfrac{15}{2}\Big)$

$\sf log~7,5=log~15-log~2$

$\sf log~7,5=log~(3\cdot5)-log~2$

$\sf log~7,5=log~3+log~5-log~2$

$\sf \red{log~7,5=y+z-x}$

d)

$\sf log~3^2=2\cdot log~3$

$\sf \red{log~3^2=2y}$