przedstaw za pomocą

[damalu]

wiedząc, że \(\displaystyle{ log_{12}4=a}\), wyznacz \(\displaystyle{ log_{12} \frac{1}{4 \sqrt{3} }}\)

[soku11]

\(\displaystyle{ \log_{12}4=a\\
\frac{\log_{2}4}{\log_{2}12}=a\\
\frac{2}{\log_{2}(3\cdot 4)}=a\\
\frac{2}{\log_{2}3+\log_{2}4}=a\\
\frac{2}{\log_{2}3+2}=a\\
2=a\log_{2}3+2a\\
a\log_{2}3=2-2a\\
\log_{2}3=\frac{2-2a}{a}\\
log_{12}\frac{1}{4 \sqrt{3} } =
-\log_{12} 4\sqrt{3}=
-\frac{\log_{2}4\sqrt{3}}{\log_{2}12}=
-\frac{\log_{2}4+\log_{2}\sqrt{3}}{\log_{2}(3\cdot 4)}=
-\frac{2+\frac{1}{2}\log_{2}3}{\log_{2}3+\log_{2}4}=
-\frac{2+\frac{1}{2}\log_{2}3}{\log_{2}3+2}=
-\frac{2+\frac{1}{2}\frac{2-2a}{a}}{2+\frac{2-2a}{a}}}\)

Pozdrawiam.

[miki999]

Albo tak:

\(\displaystyle{ log _{12} \ \frac{1}{4 \sqrt{3} } = \frac{1}{2} log _{12} \ \frac{1}{16*3} =- \frac{1}{2} log _{12}\ 48=- \frac{1}{2} *[log _{12}\ 12+log _{12}\ 4]= -\frac{1}{2}- \frac{1}{2}a}\)