Question

Calcule o valor de x nos seguintes logaritmos.


Alguém pode me ajudar? ​

Answer 1

Aplicando a definição dos logaritmos

㏒ₐ b = n  ⇔   aⁿ = b

Sendo assim

㏒₄ x = 3 ⇔ 4³ = x ⇒ x = 64

㏒₅ x = 2 ⇔ 5² = x ⇒ x = 25

㏒₂ 16 = x ⇔ 2ˣ = 16 ⇒ 2ˣ = 2⁴  dai   x = 4

16^(1/4) = 2 entao

16^(1/4) = ㏒₃ x ⇔ 2 = ㏒₃ x ⇒ 3² = x ⇒ x = 9

㏒ₓ 125 = 3 ⇔ x³ = 125 ⇒ x = 5

Answer 2

$\large\boxed{\begin{array}{l}\underline{\rm D~\!\!efinic_{\!\!,}\tilde ao~de~logaritmo}\\\sf\ell og_ba=x\Longleftrightarrow b^x=a\begin{cases}\sf a>0\\\sf b>0\\\sf b\ne1\end{cases}\\\tt a)~\sf\ell og_4x=3\\\sf x=4^3=64\\\tt b)~\sf\ell og_5x=2\\\sf x=5^2=25\\\tt c)~\sf\ell og_216=x\\\sf 2^x=16\\\sf 2^x=2^4\\\sf x=4\\\tt d)~\sf16^{\frac{1}{4}}=\ell og_3x\\\sf\ell og_3x=(2^4)^{\frac{1}{4}}\\\sf\ell og_3x=2\\\sf x=3^2\\\sf x=9\\\tt e)~\sf\ell og_x125=3\\\sf x^3=125\\\sf x=\sqrt[\sf3]{\sf125}\\\sf x=5\end{array}}$