Question

$( frac{ sqrt[4]{2} }{ sqrt{3} })^{x^2+4}= 20,25^{x+1}$
Полное решение, пожалуйста

Answer 1

$(frac{sqrt[4]{2}}{sqrt3})^{x^2+4}=20,25^{x+1}\\20,25=20frac{1}{4}=frac{81}{4}=frac{3^4}{2^2}=frac{(sqrt3)^8}{(sqrt[4]{2})^8}=(frac{sqrt3}{sqrt[4]{2}})^8=(frac{sqrt[4]{2}}{sqrt{3}})^{-8}\\(frac{sqrt[4]{2}}{sqrt3})^{x^2+4}=(frac{sqrt[4]{2}}{sqrt3})^{-8(x+1)}\\x^2+4=-8x-8\\x^2+8x+12=0\\x_1=-6,; x_2=-2; (teor.; Vieta)$