Alg9.4.28

Solve $a^{3-x}\cdot b^{5x}=a^{3x}\cdot b^{x+5}\,$

Applying logs on both sides,

$\log (a^{3-x})+\log(b^{5x})=\log(a^{3x}+\log(b^{x+5})\,$

Simplifying

$(3-x)\log a+5x\log b=3x\log a+(x+5)\log b\,$

$3\log a-x\log a+5x\log b=3x\log a+x\log b+5\log b\,$

$3\log a-5\log b=4x\log a -4x\log b\,$

$3\log a-5\log b=4x(\log a-\log b)\,$

$\log a^3-\log b^3=4x(\log a-\log b)\,$

Hence

$x=\frac{\log a^3-\log b^3}{\log a^4-\log b^4}\,$

$x=\frac{\log(\frac{a^3}{b^3})}{\log(\frac{a^4}{b^4})}\,$