Question

sabendo que log3= 0,477, log 5= 0,699 e log7= 0,845 calcule o logaritmo de log3 49


considere a função F: r - r definida por Log10 (x² -6x -6). calcule f (-2).​

Answer 1

$\large\boxed{\begin{array}{l}\sf 1)\,Sabendo\,que\,\ell og3=0,477,\ell og5=0,699~e~\ell og7=0,845\\\sf Calcule\,o\,logaritmo\,de\,\ell og_349.\\\underline{\sf soluc_{\!\!,}\tilde ao\!:}\\\rm \ell og_349=\dfrac{\ell og49}{\ell og3}=\dfrac{\ell og7^2}{\ell og3}\\\\\rm \ell og_349=\dfrac{2\ell og7}{\ell og3}=\dfrac{2\cdot0,845}{0,477}\\\\\rm \ell og_349=\dfrac{1,69}{0,477}\\\\\rm \ell og_349=3,54\end{array}}$

$\large\boxed{\begin{array}{l}\sf 2)\,Considere\,a\,func_{\!\!,}\tilde ao\, F:\mathbb{R}\longrightarrow\mathbb{R}\,d~\!\!efinida\,por\\\sf f(x)=\ell og_{10}x^2-6x-6. Calcule\,f(-2).\\\underline{\sf soluc_{\!\!,}\tilde ao\!:}\\\rm f(-2)=\ell og_{10}(-2)^2-6\cdot(-2)-6\\\rm f(-2)=\ell og_{10}4+12-6\\\rm f(-2)=\ell og_{10}10\\\rm f(-2)=1\end{array}}$