a)
log_53 oraz \frac{1}{log_35}
log_53=\frac{1}{log_35} bo log_ab=\frac{1}{log_ba} wzór
b)
log_23\cdot log_34 oraz 2
log_23\cdot log_34=\frac{log_33}{log_32}\cdot log_32^2=\frac{1}{log_32}\cdot 2log_32=1\cdot 2=2
c)
\frac{3}{log_210} oraz \frac{1}{2}log4+\frac{2}{3}log8
\frac{1}{2}log4+\frac{2}{3}log_8=log4^{\frac{1}{2}}+log(2^3)^{\frac{2}{3}}=log\sqrt4+log2^2=log2+log4=log(2\cdot4)=
=log8=log2^3=3log2=3\cdot log_{10}2=3\cdot \frac{1}{log_210}=\frac{3}{log_210}
d)
log96^{0,25}-\frac{1}{4}log\frac{2}{27} oraz \frac{1}{log_610}
{log96^{0,25}-\frac{1}{4}log\frac{2}{27}=log96^{\frac{1}{4}}-log(\frac{2}{27})^{\frac{1}{4}}=log96^{\frac{1}{4}}:(\frac{2}{27})^{\frac{1}{4}}=log(96:\frac{2}{27})^{\frac{1}{4}}=}
=log(\not96^{48}\cdot \frac{27}{\not2^1})^{\frac{1}{4}}=log\sqrt[4]{48\cdot 27}=log\sqrt[4]{1296}=log6=log_{10}6=\frac{1}{log_610}
Zastosowane wzory
log_aa=1
log_ab=\frac{1}{log_ba}
log_ax+log_ay=log_a(x\cdot y)
log_ax-log_ay=log_a\frac{x}{y}