Równania log z zmianą podstaw

winfast29 · Post autor: **winfast29** » 7 gru 2008, o 07:03

\(\displaystyle{ 4^x 3^{x-2} =2^x}\)

\(\displaystyle{ log_{0,5}x log _{4}x = -4}\)

Nakahed90 · Post autor: **Nakahed90** » 7 gru 2008, o 08:35

\(\displaystyle{ 4^x * 3^{x-2} =2^x \ \ \ D:x\in R}\)
\(\displaystyle{ (\frac{4}{2})^{x}*\frac{3^{x}}{9}=1}\)
\(\displaystyle{ 2^{x}*3^{x}=9}\)
\(\displaystyle{ 6^{x}=9}\)
\(\displaystyle{ x=log_{6}9}\)

winfast29 · Post autor: **winfast29** » 7 gru 2008, o 13:36

a drugi przyklad;-)

Nakahed90 · Post autor: **Nakahed90** » 7 gru 2008, o 13:55

\(\displaystyle{ \frac{log_{2}x}{log_{2}\frac{1}{2}}*\frac{log_{2}x}{log_{2}4}=-4 \ \ D:x>0}\)
\(\displaystyle{ \frac{log_{2}x}{-1}*\frac{log_{2}x}{2}=-4}\)
\(\displaystyle{ log^{2}_{2}x=8}\)
\(\displaystyle{ log_{2}x=2\sqrt{2} log_{2}x=-2\sqrt{2}}\)
\(\displaystyle{ x=2^{2\sqrt{2}} x=(\frac{1}{2})^{2\sqrt{2}}}\)