Question

Calcule logx 20, sendo logx 2 = a, logx 5 = b.

Answer 1

Resposta:

\begin{gathered}log_x \sqrt[3]{12} \\ \\ log_x12^{ \frac{1}{3} } \\ \\ \frac{1}{3}.log_x12 \\ \\ \frac{1}{3}.log_x(2.2.3) \\ \\ \frac{1}{3}.(log_x2+log_x2+log_x3) \end{gathered}

log

x

3

12

log

x

12

3

1

3

1

.log

x

12

3

1

.log

x

(2.2.3)

3

1

.(log

x

2+log

x

2+log

x

3)

como:

\begin{gathered}log_x2=a \\ \\ log_x3=b\end{gathered}

log

x

2=a

log

x

3=b

substituindo:

\begin{gathered} \frac{1}{3}.(log_x2+log_x2+log_x3) \\ \\ \frac{1}{3}.(a+a+b) \\ \\ \frac{1}{3}.(2a+b) \\ \\ \frac{2a}{3}+ \frac{b}{3} \\ \\ \frac{2a+b}{3} } \end{gathered}

lembrando que 2.2.3=12 eu apenas substituir isso pra pode aparecer as relaçoes que eu quero.