Question

qual e o valor da expressão: Log2 0,25 + log3 V27 + Log4 64

Answer 1

Resolução da questão, veja:

Vamos determinar cada um dos logaritmos, para depois se apegarmos a expressão, veja:

$\mathsf{log_{2}}^~\mathsf{\frac{1}{4}}}\\\\\\ \mathsf{2^{x}} = \mathsf{\dfrac{1}{4}}}\\\\\\ \mathsf{2^{x}} = \mathsf{\dfrac{1}{2^{2}}}}\\\\\\ \mathsf{\diagup\!\!\!\!2^{x}}} = \mathsf{\diagup\!\!\!\!2^{-2}}}\\\\\\ \large\boxed{\boxed{\boxed{\boxed{\boxed{\mathsf{x = -2.}}}}}}}}}}}}}$

Agora vamos calcular o segundo logaritmo:

$\mathsf{log_{3}^~{\sqrt{27}}}}\\\\\\ \mathsf{3^{x}}}} = \mathsf{27^{\frac{1}{2}}}}}}\\\\\\ \mathsf{3^{x} = 3^{3}^{^{\frac{1}{2}}}}}}}\\\\\\\ \mathsf{\diagup\!\!\!\!3^{x} = \diagup\!\!\!\!3^{\frac{3}{2}}}}}\\\\\\\\ \large\boxed{\boxed{\boxed{\boxed{\boxed{\mathsf{x = \dfrac{3}{2}}}}}}}}}}}}}}}}$

Agora vamos calcular o terceiro logaritmo, veja:

$\mathsf{log_{4}^~{64}}}\\\\\\\ \mathsf{2^{2}^{x}}} = \mathsf{2^{6}}}\\\\\\\ \mathsf{ \diagup\!\!\!\!2^{2x}}}} = \mathsf{ \diagup\!\!\!\!2^{6}}}\\\\\\\ \mathsf{2x = 6}}\\\\\\ \mathsf{x = \dfrac{6}{2}}\\\\\\\ \large\boxed{\boxed{\boxed{\boxed{\boxed{\mathsf{x = 3.}}}}}}}}}}}}}}}}}}}}}}}}}}$

Agora que temos os 3 logaritmos podemos então fazer a soma dos mesmos, veja:

$\mathsf{-2 + \dfrac{3}{2} + 3}}\\\\\\\ \mathsf{\dfrac{-4+3+6}{2}}}\\\\\\\ \large\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\mathsf{\dfrac{5}{2}}}}}}}}}}}}}}}}}}$

Ou seja, a soma destes logaritmos é igual a Cinco meios.

Espero que te ajude '-'