Korzystamy z definicji logarytmu:
[tex]\bold{\log_ab=x\quad\iff\quad a^x = b}[/tex]
oraz działań na potęgach.
[tex]\log_{\sqrt[3]6}6\sqrt6=x\quad\iff\quad (\sqrt[3]6)^x = 6\sqrt6\\\\\\\left(\big6^\frac13\right)^x = 6\cdot\big6^\frac12\\\\ \big6^{\frac13\cdot x} = \big6^{1+\frac12}\\\\ \big6^{\frac13 x} = \big6^{\frac32}\\\\\frac13 x=\frac32\qquad/\cdot3\\\\x=\frac92=4,5\\\\\\\log_{\sqrt[3]6}6\sqrt6=4,5[/tex]
[tex]\log_{\sqrt[3]{10}}\sqrt{10^5}=x\quad\iff\quad (\sqrt[3]{10})^x = \sqrt{10^5}\\\\\\\left(1\big0^\frac13\right)^x =\left(1\big0^5\right)^\frac12\\\\ 1\big0^{\frac13\cdot x} = 1\big0^{5\cdot\frac12}\\\\ 1\big0^{\frac13x} = 1\big0^{\frac52}\\\\\frac13 x=\frac52\qquad/\cdot3\\\\x=\frac{15}2=7,5\\\\\\ \log_{\sqrt[3]{10}}\sqrt{10^5}=7,5[/tex]
[tex]\log_{\sqrt7}\sqrt[3]{49}=x\quad\iff\quad \left(\sqrt7\right)^x = \sqrt[3]{49}}\\\\\\\left(\big7^\frac12\right)^x =\left(\big7^2\right)^\frac13\\\\ \big7^{\frac12x}=\big7^\frac23\\\\\frac12x=\frac23\qquad/\cdot2\\\\x=\frac43=1,(3)\\\\\\\log_{\sqrt7}\sqrt[3]{49}=1,(3)[/tex]
