Question

به
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​

Answer 1

Resposta:

a) log, 27 = x 3^x = 27 3^x = 33 x = 3

b) logs 125 = x 5^x = 125 5^x = 53 x = 3

c) log 10000 = x 10^x = 10000 10^x = 104 X = 4

d) log1)2 32 = X (1/2)^x = 32 1/2^x = 25 2^(-x) = 25 -X = 5 x = -5

e) log 0.01 = x 10 ^ x = 0.01 10 ^ x = 10-2 X = -2

f) log, 0.5 = x 2 ^ x = 0.5 2 ^ x = 1/2 2 ^ x = 2- X = -1

g) log2 v8 = x 2 ^ x = v8 2 ^ x = v23 2 ^ x = 2 ^ (3/2) X = 3/2 h) log4 v32 = x 4 ^ x = v32 4 ^ x = v25 (2) ^ x = 2 ^ (5/2) 2 ^ 2x = 2 ^ (5/2)2x = 5/2 x = 5/4

Vou usar esse símbolo "A" número está elevado a outro! para indicar que um a) x = log25 da base 5 >> 5^x 25 >> 5^x = 52 >> %3D X= 2 b) 3 = log27 da base x >> x^3 = 27 >> x^3 =33 >> X= 3 c) x = log7 da base 7>> 7^x = 7 >> x= 1 d) x = log1/25 da base 5 >> 5^x = 1/25 >> 5^x = 25^-1 >> 5^x = (52)^-1 >> 5^x = 5^-2 >> x= -2

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