Question

1) Usando a definição de logaritmo, calcule:​

Answer 1

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

$\large\boxed{\begin{array}{l}\rm 1)~Usando~a~d\!\,efinic_{\!\!,}\tilde ao~de~logaritmo,calcule:\\\tt a)~\rm \ell og_327\\\tt b)~\rm\ell og_5125\\\tt c)~\rm\ell og_20,5\\\tt d)~\rm\ell og_4\sqrt{32}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm soluc_{\!\!,}\tilde ao\!:}\\\tt a)~\sf\ell og_327=x\\\sf 3^x=27\\\sf 3^x=3^3\\\sf x=3\\\tt b)~\sf\ell og_5125=x\\\sf 5^x=125\\\sf 5^x=5^3\\\sf x=3\\\tt c)~\sf\ell og_20,5=x\\\sf 2^x=0,5\\\sf 2^x=\dfrac{5}{10}\\\sf 2^x=\dfrac{1}{2}\\\sf 2^x=2^{-1}\\\sf x=-1\\\tt d)~\sf \ell og_4\sqrt{32}=x\\\sf 4^x=\sqrt{32}\\\sf (2^2)^x=\sqrt{2^5}\\\sf 2^{2x}=2^{\frac{5}{2}}\\\sf 2x=\dfrac{5}{2}\\\sf 4x=5\\\sf x=\dfrac{5}{4}\end{array}}$

$\large\boxed{\begin{array}{l}\rm 2)~Determine~o~valor~de~x~em~cada~caso:\\\tt a)~\rm\ell og_264=x\\\tt b)~\rm\ell og_x216=3\\\tt c)~\rm \ell ogx=0\\\tt d)~\rm 2=\ell og_x625\end{array}}$

$\large\boxed{\begin{array}{l}\rm\underline{\rm soluc_{\!\!,}\tilde ao}\\\tt a)~\sf\ell og_264=x\\\sf 2^x=64\\\sf 2^x=2^6\\\sf x=6\\\tt b)~\sf\ell og_x216=3\\\sf x^3=216\\\sf x=\sqrt[\sf3]{\sf 216}\\\sf x=6\\\tt c)~\sf \ell og x=0\\\sf x=10^0\\\sf x=1\\\tt d)~\sf 2=\ell og_x625\\\sf x^2=625\\\sf x=\sqrt{625}\\\sf x=25\end{array}}$

$\large\boxed{\begin{array}{l}\rm 3)~Classifique~em~falso~ou~verdadeiro:\\\tt a)~\rm \ell og_51=1\\\tt b)~\rm\ell og_15=5\\\tt c)~\rm\ell og_55=1\\\tt d)~\rm\ell og_51=0\\\tt e)~\rm \ell og_73^7=3\\\tt f)~\rm\ell og_33^7=7\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm soluc_{\!\!,}\tilde ao\!:}\\\tt a)~\sf falso~pois~\ell og_51=0\\\tt b)~\sf falso~pois\,\not\exists\,\,\ell og_15~\\\tt c)~\sf verdadeiro~pois~\ell og_bb=1~\forall,b\in\mathbb{N}\\\tt d)~\sf verdadeiro~pois~\ell og_b1=0~\forall b\ne1~e~b>0\\\tt e)~\sf falso~pois 7^3\ne3^7\\\tt f)~\sf verdadeiro~pois~ 3^7=3^7\end{array}}$