Ответ:

55) Пользуемся свойствами:  [tex]log_{a^{k}}\, b=\dfrac{1}{k}\cdot log_{a}\, b\ \ ,\ \ log_{a}\, b^{k}=k\cdot log_{a}\, b[/tex]  .

[tex]log_{a}\dfrac{b}{c}=log_{a}\, b-log_{a}\, c\ \ ,\ \ \ a > 0\ ,\ a\ne 1\ ,\ b > 0\ ,\ c > 0[/tex]  .

[tex]log_{\sqrt[4]{7}}169=a\ \ \ \Rightarrow \ \ \ log_{7^{\frac{1}{4}}}13^2=\dfrac{2}{1/4}\cdot log_{7}\, 13=a\ \ ,\ \ log_7\, 13=\dfrac{a}{8}\ ;[/tex]  

[tex]log_{7\sqrt7}\, \dfrac{49}{13}=log_{7^{\frac{3}{2}}}\, \dfrac{7^2}{13}=\dfrac{2}{3}\cdot log_7\, \dfrac{7^2}{13} =\dfrac{2}{3}\cdot \Big(log_7\, 7^2-log_7\, 13\Big)=\\\\\\=\dfrac{2}{3}\cdot \Big(2\, \underbrace{log_77}_{1}-log_7\, 13\Big)=\dfrac{2}{3}\cdot \Big(2-\dfrac{a}{8}\Big)=\dfrac{16-a}{12}[/tex]    

56) Решаем аналогично.

[tex]log_{\sqrt[4]{2}}\sqrt{7}=a\ \ \Rightarrow \ \ \ log_{2^{\frac{1}{4}}}\, 7^{\frac{1}{2}}=\dfrac{4}{2}\cdot log_2\, 7=a\ \ ,\ \ log_2\, 7=\dfrac{a}{2}\ ;[/tex]

[tex]log_{2\sqrt[3]{2}}\, \dfrac{7}{16}=log_{2^{\frac{4}{3}}}\dfrac{7}{2^4}=\dfrac{3}{4}\cdot log_2\, \dfrac{7}{2^4}=\dfrac{3}{4}\cdot \Big(log_27-log_2\, 2^4\Big)=\\\\\\=\dfrac{3}{4}\cdot \Big(\dfrac{a}{2}-4\cdot log_22\Big)=\dfrac{3}{4}\cdot \Big(\dfrac{a}{2}-4\Big)=\dfrac{3\, (a-8)}{8}[/tex]