Ответ:

Применяем свойства логарифмов.

[tex]\bf 1)\ \ log_3\, 2=a\ ,\ \ \ log_{10}\, 3=b\\\\log_6\, 0,3=log_6\dfrac{3}{10}=log_6 3-log_6\, 10=\dfrac{1}{log_3\, 6}-\dfrac{1}{log_{10}\, 6}=\\\\\\=\dfrac{1}{log_3\, (2\cdot 3)}-\dfrac{1}{log_{10}\, (2\cdot 3)}=\dfrac{1}{log_3\, 2+log_3\, 3}-\dfrac{1}{log_{10}\, 2+log_{10}\, 3}=\\\\\\=\dfrac{1}{log_3\, 2+1}-\dfrac{1}{log_{10}\, 2+b}=\dfrac{1}{a+1}-\dfrac{1}{\dfrac{log_3\, 2}{log_3\, 10}+b}=\\\\\\=\dfrac{1}{a+1}-\dfrac{1}{log_3\, 2\cdot log_{10}\, 3+b}=\dfrac{1}{a+1}-\dfrac{1}{a\cdot b+b}=[/tex]

[tex]\bf =\dfrac{1}{a+1}-\dfrac{1}{b\cdot (a+1)}=\dfrac{b-1}{b\, (a+1)}[/tex]  

[tex]\bf 2)\ \ log_{10}\, 2=a\ \ ,\ \ log_{10}\, 7=b\\\\log_8\, 98=log_{2^3}\, (2\cdot 7^2)=\dfrac{1}{3}\cdot log_2\, (2\cdot 7^2)=\dfrac{1}{3}\cdot \Big(log_2\, 2+log_2\, 7^2\Big)=\\\\\\=\dfrac{1}{3}\cdot \Big(1+2\cdot log_2\, 7\Big)=\dfrac{1}{3}\cdot \Big(1+2\cdot \dfrac{log_{10}\, 7}{lg_{10}\, 2}\Big)=\dfrac{1}{3}\cdot \Big(1+2\cdot \dfrac{b}{a}\Big)=\\\\\\=\dfrac{1}{3}\cdot \dfrac{a+2b}{a}=\dfrac{a+2b}{3a}[/tex]