Question

4.(UFRGS/2008) A solução da equação (0.01)
- 50 €
a) -1 + log12
b) 1 + log2
0) 1+ log2
d) 1. log2
e) 2log2​

Answer 1

Resposta:

0,01ˣ =50

log 0,01ˣ = log 50

x * log 0,01 = log (5²*2)

x * log 10⁻²  = log 5² + log 2

*******log 5 = log 10/2  = log 10 -log 2 =1-log 2

-2x * log 10 = 2 * log 5 + log 2

-2x * 1 = 2 * (1-log 2) + log 2

-2x = 2 -2log 2 + log 2

-2x = 2- log 2

x =(1/2) * log 2  -1

x=  log √2   -1   = -1   + log√2

Letra A

Answer 2

• A solução dessa equação exponencial é igual a -1+log√2. ( A ).

Para resolver essa equação exponencial, devemos deixar as bases iguais.

$\large\displaystyle\begin{aligned} (0,01)^x = 50 \end{aligned}$

$\large\displaystyle\begin{aligned} \left(\frac{1}{100} \right)^x = 50 \end{aligned}$

$\large\displaystyle\begin{aligned} \left(\frac{1}{10^2} \right)^x = 50 \end{aligned}$

$\large\displaystyle\begin{aligned} \left(10^{-2} \right)^x = 50 \end{aligned}$

Perceba que não podemos igualar as bases para cortar. Então, aplique o logaritmo natural nos dois lados. Logo:

$\large\displaystyle\begin{aligned} \left(10\right)^{-2x} = 50 \end{aligned}$

$\large\displaystyle\begin{aligned} \log\left(10\right)^{-2x} = \log(50 )\end{aligned}$

$\large\displaystyle\begin{aligned} -2x\cdot \log\left(10\right) = \log(5\cdot 10 )\end{aligned}$

$\large\displaystyle\begin{aligned} -2x\cdot 1 = \log(5) + \log ( 10 )\end{aligned}$

$\large\displaystyle\begin{aligned} -2x= \log(5) + 1\end{aligned}$

Lembrando que o log de 5 é igual ao $\large\displaystyle\begin{aligned} \log(5) = \log \left( \frac{10}{2} \right)\end{aligned}$ , sabendo que o log da divisão é a subtração dos logs. Logo: $\large\displaystyle\begin{aligned} \log(5) = \log (10) - \log (2) \end{aligned}$ . Ficando assim: $\large\displaystyle\begin{aligned} \log(5) = 1 - \log (2) \end{aligned}$.

$\large\displaystyle\begin{aligned} -2x= \log(5) + 1\end{aligned}$

$\large\displaystyle\begin{aligned} -2x= 1-\log(2) + 1\end{aligned}$

$\large\displaystyle\begin{aligned} -2x= 2-\log(2)\ \ \cdot \ \ (-1) \end{aligned}$

$\large\displaystyle\begin{aligned} 2x= -2+\log(2) \end{aligned}$

$\large\displaystyle\begin{aligned} x= \frac{-2}{2}+\frac{\log(2)}{2} \end{aligned}$

$\large\displaystyle\begin{aligned} x= -1+\frac{1}{2} \cdot \log(2) \end{aligned}$

Temos agora que aplicar aquela propriedade do peteleco. Só que voltando para o log. Ficando:

$\large\displaystyle\begin{aligned} x= -1+\frac{1}{2} \cdot \log(2) \end{aligned}$

$\large\displaystyle\text{ \begin{aligned} x= -1+ \log(2)^{\frac{1}{2}} \end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} \therefore \boxed{\boxed{\green{x= -1+ \log\sqrt{(2)}}}}\end{aligned} }$

Veja mais sobre:

Logaritmo.

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