Question

o valor de x na equação exponencial está indicada corretamente no item:

(questão 4 e 5)

Answer 1

$\large\boxed{\begin{array}{l}\sf 4)~O\,valor\,de\,x\,na\,equac_{\!\!,}\tilde ao\,exponencial \,64^{x^2-2x}=1\\\sf est\acute a\,indicado\,corretamente\,no\,item\\\sf a-\,V=\{1,0\}\\\sf\!\!\bullet b-\,V=\{2,0\}\\\sf c-\,V=\{-1,\!-2\}\\\sf d-\, V=\{-1,0\}\\\sf e-\, V=\{-2,0\}\\\underline{\sf soluc_{\!\!,}\tilde ao\!:}\\\rm 64^{x^2-2x}=1\\\rm 64^{x^2-2x}=64^0\\\rm x^2-2x=0\\\rm x\cdot(x-2)=0\\\rm x=0\\\rm x-2=0\\\rm x=2\\\rm V=\{2,0\}\end{array}}$

$\large\boxed{\begin{array}{l}\sf 5)~ O\,valor\,de\,x\,na\,equac_{\!\!,}\tilde ao\,exponencial\,144^{x^2-5x+6}=1\\\sf est\acute a\,indicado\,corretamente\,no\,item\\\sf\!\!\! \bullet a- V=\{2,3\}\\\sf b- V=\{2,0\}\\\rm c- V=\{2,\!-3\}\\\sf d- V=\{-2,0\}\\\sf e- V=\{-2,3\}\\\underline{\sf soluc_{\!\!,}\tilde ao\!:}\\\rm 144^{x^2-5x+6}=1\\\rm 144^{x^2-5x+6}=144^0\\\rm x^2-5x+6=0\end{array}}$

$\large\boxed{\begin{array}{l}\rm x^2-5x+6=0\\\rm\Delta=b^2-4ac\\\rm\Delta=(-5)^2-4\cdot1\cdot6\\\rm\Delta=25-24\\\rm\Delta=1\\\rm x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\rm x=\dfrac{-(-5)\pm\sqrt{1}}{2\cdot1}\\\\\rm x=\dfrac{5\pm1}{2}\begin{cases}\rm x_1=\dfrac{5+1}{2}=\dfrac{6}{2}=3\\\\\rm x_2=\dfrac{5-1}{2}=\dfrac{4}{2}=2\end{cases}\\\rm V=\{2,3\}\end{array}}$