Question

1- Determine o conjunto verdade do exponencial: 7 x = 343 2- O conjunto solução da equação é? 49x = √7 3- O valor de log8 3√16 é: 4- Qual o logaritmo de: log2 32 = x 5- Determine o valor de log 30, sabendo que log 2= 5; log 3 = 10 e log 5 =25.

Answer 1

Resposta:

$\text{Leia abaixo}$

Explicação passo-a-passo:

1.

$7^x = 343$

$7^x = 7^3$

$\boxed{\boxed{x = 3}}$

2.

$49^x = \sqrt{7}$

$7^{2x} = 7^{\frac{1}{2}}$

$2x = \dfrac{1}{2}$

$\boxed{\boxed{x = \dfrac{1}{4}}}$

3.

$\text{log}_8\: \sqrt[3]{16}$

$\text{log}_8\: 16^{\frac{1}{3}}$

$\text{log}_8\: 2^{\frac{4}{3}}$

$8^x = 2^{\frac{4}{3}}$

$2^{3x} = 2^{\frac{4}{3}}$

$3x = \dfrac{4}{3}$

$\boxed{\boxed{x = \dfrac{4}{9}}}$

4.

$\text{log}_2\:32 = x$

$2^x = 32$

$2^x = 2^5$

$\boxed{\boxed{x = 5}}$

5.

$\text{log } 30 = \text{log } 2.3.5 = \text{log } 2 + \text{log } 3 + \text{log } 5 = 5 + 10 + 25 = 40$

Answer 2

Explicação passo-a-passo:

1)

$\sf 7^x=343$

$\sf 7^x=7^3$

Igualando os expoentes:

$\sf \red{x=3}$

2)

$\sf 49^x=\sqrt{7}$

$\sf (7^2)^x=7^{\frac{1}{2}}$

$\sf 7^{2x}=7^{\frac{1}{2}}$

Igualando os expoentes:

$\sf 2x=\dfrac{1}{2}$

$\sf x=\dfrac{\frac{1}{2}}{2}$

$\sf x=\dfrac{1}{2}\cdot\dfrac{1}{2}$

$\sf \red{x=\dfrac{1}{4}}$

3)

$\sf log_{8}~\sqrt[3]{16}$

$\sf =log_{2^3}~\sqrt[3]{2^4}$

$\sf =log_{2^3}~2^{\frac{4}{3}}$

Lembre-se que $\sf log_{b^m}~a^n=\dfrac{n}{m}\cdot log_{b}~a$

Assim:

$\sf log_{2^3}~2^{\frac{4}{3}}$

$\sf =\dfrac{\frac{4}{3}}{3}\cdot log_{2}~2$

$\sf =\dfrac{4}{3}\cdot\dfrac{1}{3}\cdot1$

$\sf =\red{\dfrac{4}{9}}$

4)

$\sf log_{2}~32=x$

$\sf 2^x=32$

$\sf 2^x=2^5$

Igualando os expoentes:

$\sf \red{x=5}$

5)

$\sf log~30=log~(2\cdot3\cdot 5)$

$\sf log~30=log~2+log~3+log~5$

$\sf log~30=5+10+25$

$\sf \red{log~30=40}$