Question

1- Usando
a definição, calcule:
logaritimando 25.

B) a base, dados logaritmo igual a 3
e o logaritmando 27.

c) O logaritmo de 7 ma base 7

D) o logaritmo de 1 sobre 25 na base 5

por favor me ajude é urgentee​

Answer 1

Resposta:

$b) \: log_{a}(27) = 3\\ log_{a}( 27) = {3}^{3} \\ a = 3$

$c) \: log_{7}(7) = x \\ 7 {}^{x} = 7 \\ x = 1$

$d) \: log_{5}( \frac{1}{25} ) = x \\ 5 {}^{x} = \frac{1}{25} \\ {5}^{x} = \frac{1}{ {5}^{2} } \\ {5}^{x} = 5 {}^{ - 2} \\ x = - 2$

Answer 2

Oie, Td Bom?!

a)

$log_{10}(25) = x$

$log_{10}(5 {}^{2} ) = x$

• $log_{a}(b {}^{c} ) = c \: . \: log_{a}(b)$.

$2 log_{10}(5) = x$

$x = 2 log_{10}(5)$

$x≈1,39794...$

b)

$log_{x}(27) = 3$

• $log_{a}(x) = b⇒x = a {}^{b}$.

$27 = x {}^{3}$

$x {}^{3} = 27$

$x {}^{3} = 3 {}^{3}$

$x = \sqrt[3]{3 {}^{3} }$

$x = 3$

c)

$log_{7}(x) = 7$

• $log_{a}(x) = b⇒x = a {}^{b}$.

$x = 7 {}^{7}$

$x = 823.543$

d)

$log_{5}( \frac{1}{25} ) = x$

$log_{5}(5 {}^{ - 2} ) = x$

• $log_{a}(a {}^{x} ) = x \: . \: log_{a}(a)$.

$- 2 log_{5}(5) = x$

$- 2 log_{5}(5) = x$

$- 2 \: . \: 1 = x$

$- 2 = x$

$x = - 2$

Att. Makaveli1996