Question

como resolve isso? obg

Answer 1

$\log_4(-x^2+10x-15)=0 \\-x^2+10x-15 = 1\\x^2-10x+16=0\\(x-8)(x-2)=0\\\\x\in\{2,8\}\longleftarrow essas~s\~ao ~as~ra\'ices$

Portanto a soma é 2+8 = 10

Answer 2

$\boxed{\begin{array}{l}\underline{\rm D~\!\!efinic_{\!\!,}\tilde ao~de~logaritmo}\\\sf \ell og_ba=x\Longleftrightarrow b^x=a~com~\begin{cases}\sf a>0\\\sf b>0\\\sf b\ne1\end{cases}\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm condic_{\!\!,}\tilde ao~de~exist\hat encia\!:}\\\sf -x^2+10x-15>0\\\sf fazendo~g(x)=-x^2+10x-15\\\underline{\rm ra\acute izes:}\\\sf -x^2+10x-15=0\cdot(-1)\\\sf x^2-10x+15=0\\\sf x^2-10x+25-10=0\\\sf (x-5)^2-10=0\\\sf (x-5)^2=10\\\sf x-5=\pm\sqrt{10}\\\sf x-5=\sqrt{10}\\\sf x=5+\sqrt{10}\\\sf x-5=-\sqrt{10}\\\sf x=5-\sqrt{10}\\\sf g(x)>0\implies 5-\sqrt{10}<x<5+\sqrt{10}\end{array}}$

$\boxed{\begin{array}{l}\sf -x^2+10x-15=4^0\\\sf -x^2+10x-15=1\\\sf -x^2+10x-15-1=0\\\sf -x^2+10x-16=0\cdot(-1)\\\sf x^2-10x+16=0\\\sf x^2-10x+25-9=0\\\sf (x-5)^2-9=0\\\sf (x-5)^2=9\\\sf x-5=\pm\sqrt{9}\\\sf x-5=\pm3\\\sf x-5=3\\\sf x=3+5\\\sf x=8\\\sf x-5=-3\\\sf x=5-3\\\sf x=2\end{array}}$

$\boxed{\begin{array}{l}\sf como~as~ra\acute izes~pertencem~a~condic_{\!\!,}\tilde ao~de~exist\hat encia\\\sf ent\tilde ao\\\sf x_1=8~e~x_2=2\\\sf a~soma~\acute e~dada~por~8+2=10\end{array}}$