Question

Como resolver: log 2√2 na base 1/4

Answer 1

Resolvendo a primeira:

\begin{gathered}log _{ \frac{1}{4}} 2 \sqrt{2} = x \\ \\ ( 1/4)^{x} = 2 \sqrt{2} \\ (1/ 2^{2})^{x} = \sqrt{2* 2^{2} } \\ (2^{-2})^{x} = \sqrt{ 2^{3} } \\ (2)^{-2x} = 2^{ \frac{3}{2} }\end{gathered}

log

4

1

2

2

=x

(1/4)

x

=2

2

(1/2

2

)

x

=

2∗2

2

(2

−2

)

x

=

2

3

(2)

−2x

=2

2

3

Igualando os expoentes:

\begin{gathered}-2x = 3/2 \\ x=(3/2)/-2 \\ x= -3/4\end{gathered}

−2x=3/2

x=(3/2)/−2

x=−3/4

Resolvendo a segunda:

\begin{gathered}log _{ 2}0,25 = x \\ \\ 2^{x} = 0,25 \\ 2^{x} = 1/4 \\ 2^{x} = (1/ 2^{2}) \\ 2^{x} = 2^{-2}\end{gathered}

log

2

0,25=x

2

x

=0,25

2

x

=1/4

2

x

=(1/2

2

)

2

x

=2

−2

Igualando os expoentes:

x = -2x=−2