Oblicz (logarytmy w wykładniku)

[Abrakadabra1]

\(\displaystyle{ 4 ^{log _{2}3+ 3log_{16}2}}\)

\(\displaystyle{ 5^{2log_{5}3- log_{25}9+log_{0,2}27}}\)

[soku11]

1.
\(\displaystyle{ 4^{\log_{2}3+\log_{16}8}=
4^{\log_{2}3+\frac{\log_{2} 8}{\log_{2}16}}=
4^{\log_{2}3+\frac{3}{4}}=
4^{\log_{2}3+\frac{3}{4}\log_{2}2}=
4^{\log_{2}3+\log_{2}2^{\frac{3}{4}}}=
4^{\log_{2}(3\cdot 2^{\frac{3}{4}})}=
2^{2\log_{2}(3\cdot 2^{\frac{3}{4}})}=
2^{\log_{2}(3^2\cdot 2^{\frac{3}{2}})}=
9\cdot 2^{\frac{3}{2}}}\)

[Abrakadabra1]

Hmm mi też tak wyszło, ale w odp mam 18\(\displaystyle{ \sqrt{2}}\)

[soku11]

\(\displaystyle{ 9\cdot 2^{\frac{3}{2}}=9\cdot 2\cdot 2^{\frac{1}{2}}=18\sqrt{2}}\)

POZDRO

[ Dodano: 2 Maj 2008, 11:58 ]
\(\displaystyle{ 5^{\log_{5}9- \frac{\log_{5}9}{\log_{5}25}+\frac{\log_{5}27}{\log_{5}0,2}}=
5^{\log_{5}9- \frac{1}{2}\log_{5}9- \log_{5}27}=
5^{\frac{1}{2}\log_{5}9- \log_{5}27}=
5^{\log_{5}3- \log_{5}27}=
5^{\log_{5}\frac{3}{27}}=
\frac{1}{9}}\)

POZDRO

[Abrakadabra1]

Hehe, zmyliło mnie to 2/3

Już kumam wszystko, thx.