Question

1- calcule
o valor de X:
a) log 2 x=7
b) log (1 sobre 3) ×=2
c) log x 27 = 3
d) log× 125 sobre 8 = 3
e) log 3 1 sobre 9 = ×
f) log 4 1 sobre 32 = ×

2- 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 por favor me ajudem ​

Answer 1

Resposta:

1)

a)

$a) \: log(2x) = 7 \\ 10 {}^{7} = 2x \\ 10 \: 000 \: 000 = 2x \\ x = 10 \: 000 \: 000 \div 2 \\ x = 5 \: 000 \: 000$

b)

$b) \: log( \frac{1}{3}x ) = 2 \\ {10}^{2} = \frac{1}{3} x \\ 100 = \frac{x}{3} \\ x = 300$

c)

$c) \: log(27x) = 3 \\ 10 {}^{3} = 27x \\ 1000 = 27x \\ x = \frac{1000}{27}$

d)

$d) \: log( \frac{125x}{8} ) = 3 \\ {10}^{3} = \frac{125x}{8} \\ 1000 = \frac{125x}{8} \\ 125x = 8000 \\ x = \frac{8000}{125}$

e)

$log( \frac{31}{9} ) = x \\ 10 {}^{x} = \frac{31}{9} \\ x = 0.53712$

f)

$f) \: log( \frac{41}{32} ) = x \\ 10 {}^{x} = \frac{41}{32} \\ x = 0.107634$

2)

logaritmando 25

$log(25) = x \\ 10 {}^{x} = 25 \\ x = 1.397940$

B)

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

c)

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

D)

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

Bons Estudos!

Answer 2

Oie, Td Bom?!

1. a)

$log_{2}(x) = 7$

• $log_{a}(x) = b⇒x = a {}^{b}$.

$x = 2 {}^{7}$

$x = 128$

b)

$log_{ \frac{1}{3} }(x) = 2$

• $log_{a}(x) = b⇒x = a {}^{b}$.

$x = (\frac{1}{3} ) {}^{2}$

$x = \frac{1}{9}$

c)

$log_{x}(27) = 3$

• $log_{a}(x) = b⇒x = a {}^{b}$.

$27 = x {}^{3}$

$x {}^{3} = 27$

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

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

$x = 3$

d)

$log_{x}( \frac{125}{8} ) = 3$

• $log_{a}(x) = b⇒x = a {}^{b}$.

$\frac{125}{8} = x {}^{3}$

$x {}^{3} = \frac{125}{8}$

$x {}^{3} = ( \frac{5}{2} ) {}^{3}$

$x = \sqrt[3]{( \frac{5}{2} ) {}^{3} }$

$x = \frac{5}{2}$

e)

$log_{3}( \frac{1}{9} ) = x$

$log_{3}(3 {}^{ - 2} ) = x$

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

$- 2 log_{3}(3) = x$

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

$- 2 = x$

$x = - 2$

f)

$log_{4}( \frac{1}{32} ) = x$

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

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

$\frac{ - 5}{2} \: . \: log_{2}(2) = x$

$- \frac{5}{2} \: . \: 1 = x$

$- \frac{5}{2} = x$

$x = - \frac{5}{2}$

2. 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