Ответ:

[tex]\bf log_{a}x=\dfrac{1}{5}\Big(2\, log_{a}(a-b)+3\, log_{a}(a+b)-\dfrac{1}{3}\, log_{a}a\Big)+3\, log_{a}b[/tex]  

Применим свойства логарифма :   [tex]\bf log_{a}x+log_{a}y=log_{a}(xy)\ \ ,\ \ \ log_{a}x-log_{a}y=log_{a}\dfrac{x}{y}\ \ ,\ \ k\, log_{a}x=log_{a}(x^{k})\ ,\\\\a > 0\ ,\ a\ne 1\ ,\ x > 0\ ,\ y > 0\ .[/tex]

[tex]\bf log_{a}x=\dfrac{1}{5}\Big(log_{a}(a-b)^2+log_{a}(a+b)^3-log_{a}a^{^{\frac{1}{3}}}\Big)+log_{a}b^3\\\\\\log_{a}x=\dfrac{1}{5}\, log_{a}\dfrac{(a-b)^2\cdot (a+b)^3}{b^3}+log_{a}b^3\\\\\\log_{a}x=log_{a}\Big(\dfrac{(a-b)^2\cdot (a+b)^3}{b^3}\Big)^{^{\frac{1}{5}}}+log_{a}b^3\\\\\\\log_{a}x=log_{a}\Big(\dfrac{(a^2-b^2)^{\frac{2}{5}}\cdot (a+b)^{\frac{1}{5}}}{b^{\frac{3}{5}}}\cdot b^3\Big)[/tex]      

[tex]\bf \log_{a}x=log_{a}\Big((a^2-b^2)^\frac{2}{5}\cdot (a+b)^{\frac{1}{5}}\cdot b^\frac{12}{5}}\Big)\\\\\\log_{a}x=log_{a}\sqrt[5]{\bf (a^2-b^2)^2\cdot (a+b)\cdot b^{12}}\\\\\\x=\sqrt[5]{\bf (a^2-b^2)^2\cdot (a+b)\cdot b^{12}}[/tex]