Wzory
log_ab=\frac{log_cb}{log_ca}
log_ab=\frac{1}{log_ba} \qquad (do działania c)
a)
c = 10
log_{0,1} 7=\frac{log_{10}7}{log_{10}\frac{1}{10}}=\frac{log_{10}7}{log_{10}10^{-1}}=\frac{log_{10}7}{-1}=-log_{10}7
b)
c = 2
log_8 3=\frac{log_23}{log_28}=\frac{log_23}{log_22^3}=\frac{log_23}{3}
c)
c = 49
log_7 11=\frac{log_{49}11}{log_{49}7}=\frac{log_{49}11}{log_{49}7}=\frac{log_{49}11}{\frac{1}{log_{7}49}}=\frac{log_{49}11}{\frac{1}{log_{7}7^2}}=\frac{log_{49}11}{\frac{1}{2}}=2log_{49}11
d)
c-=0,1=\frac{1}{10}
log625=log_{10}625=\frac{log_{\frac{1}{10}}625}{log_{\frac{1}{10}}10}=\frac{log_{0,1}625}{log_{\frac{1}{10}}10^{-1}}=\frac{log_{0,1}625}{-1}=-log_{0,1}625
e)
c=\frac{1}{2}
log_2 6=\frac{log_{\frac{1}{2}}6}{log_{\frac{1}{2}}2}=\frac{log_{\frac{1}{2}}6}{log_{\frac{1}{2}}(\frac{1}{2})^{-1}}=\frac{log_{\frac{1}{2}}6}{-1}=-log_{\frac{1}{2}}6
f)
c = 3
log_{\frac{1}{3}} 12=\frac{log_3 12}{log_3\frac{1}{3}}=\frac{log_312}{log_33^{-1}}=\frac{log_312}{-1}=-log_312