Question

Resolvam pfv: ∛7^x-2 = ^x√7^5

Answer 1

Verifique se eu copiei a equação certa:

$\sqrt[3]{7^{x-2}}=\sqrt[x]{7^5}  \\7^{(x-2)/3} = 7^{5/x}\\(x-2)/3=5/x\\x(x-2)=5(3)\\x^2-2x=15\\x^2-2x-15=0\\x={2+-\sqrt{(-2)^2-4(1)(-15)} }]/2\\x=(2+-\sqrt{4+60})/2\\ x' = (2+8)/2 = 10/2 = 5\\x'' = (2-8)/2 = -6/2 = -3$

Espero ter ajudado ;)

Answer 2

Resposta:

S={ -3 , 5}

Explicação:

$\sqrt[3]{7^{x-2}} =\sqrt[x]{7^5} \\ \\ Tirando~~da~~raiz\\ \\ 7^{x-2\over3}=7^{5\over x}\\ \\ cancelando~~7\\ \\ {x-2\over3}={5\over x}\\ \\ x(x-2)=3.5\\ \\ x^2-2x=15\\ \\ x^2-2x-15=0\\ \\ a=1\\ b=-2\\ c=-15\\ \\ \Delta=b^2-4ac\\ \Delta=(-2)^2-4(1)(-15)\\ \Delta=4+60\\ \Delta=64\\ \\ x={-b\pm\sqrt{\Delta} \over2a}={-(-2)\pm\sqrt{64} \over2(1)}={2\pm8\over2}\\ \\ x'={2+8\over2}={10\over2}=5\\ \\ x"={2-8\over2}-{6\over2}=-3$