Question

6. Considerando as consequências da definição dos logaritmos, calcule o valor das expressões a seguir​.

Answer 1

$\large\boxed{\begin{array}{l}\underline{\sf De\!\!~finic_{\!\!,}\tilde ao\,de\,logaritmo}\\\sf \log_ba=x\iff b^x=a\\\sf com\,a>0,b>0\,e\,b\ne1\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\sf Consequ\hat encias\,da\,de\!\!~finic_{\!\!,}\tilde ao\,de\,\log\! aritmo}\\\sf 1)~\log _b1=0.\\\underline{\sf demonstrac_{\!\!,}\tilde ao}\\\sf \log_b1=x\iff b^x=1\\\sf mas\,1=b^0\\\sf b^x=b^0\implies x=0~~\blacksquare\end{array}}$

$\large\boxed{\begin{array}{l}\sf 2)~\log_bb=1.\\\underline{\sf demonstrac_{\!\!,}\tilde ao\!:}\\\sf \log_bb=x\\\sf b^x=b\\\sf\,mas\,b=b^1.\\\sf b^x=b^1\implies x=1~\blacksquare\end{array}}$

$\large\boxed{\begin{array}{l}\sf 3)~a^{\log_ab}=b.\\\underline{\sf demonstrac_{\!\!,}\tilde ao}\\\sf fac_{\!\!,}a~\log_ab=t~e~a^{\log_ab}=x\\\sf devemos\,mostrar\,que\,x=b.\\\sf Com\,efeito, a^{\log_ab}=x\\\sf a^{t}=x\implies t=\log_ax\\\sf mas\,t=\log_ab,substituindo\,temos:\\\sf \log_ab=\log_ax\implies x=b~\blacksquare\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\sf Propriedades\,operat\acute orias\,dos\,logaritmos}\\\sf 1)~\log_c(a\cdot b)=\log_ca+\log_cb\\\underline{\sf demonstrac_{\!\!,}\tilde ao\!:}\\\sf fac_{\!\!,}a~\log_c(a\cdot b)=x,\log_ca=y\,e\,\log_cb=z\\\sf devemos\,mostrar\,que\,x=y+z.\\\sf com\,efeito,\log_c(a\cdot b)=x\implies c^x=a\cdot b~\boxed{1}\\\sf \log_ca=y\implies a=c^y~\boxed{2}\\\sf \log_cb=z\implies b=c^z~\boxed{3}\end{array}}$

$\large\boxed{\begin{array}{l}\sf substituindo\,\boxed{2}\,e\,\boxed{3}\,em\,\boxed{1}\,temos:\\\rm c^x=c^y\cdot c^z\\\sf c^x=c^{y+z}\implies x=y+z~\blacksquare\end{array}}$

$\large\boxed{\begin{array}{l}\sf 2)\log_c\bigg(\dfrac{a}{b}\bigg)=\log_ca-\log_cb\\\underline{\sf demonstrac_{\!\!,}\tilde ao\!:}\\\sf fac_{\!\!,}a~\log_c\bigg(\dfrac{a}{b}\bigg)=x,\log_ca=y~e~\log_cb=z\\\sf devemos\,mostrar\,que\,x=y-z.\\\sf Com\,efeito, \log_c\bigg(\dfrac{a}{b}\bigg)=x\implies c^x=\dfrac{a}{b}\,\boxed{1}\\\\\sf\log_ca=y\implies c^y=a\,\boxed{2}\\\sf \log_cb=z\implies c^z=b\,\boxed{3}\end{array}}$

$\large\boxed{\begin{array}{l}\sf substituindo\,\boxed{2}\,e\,\boxed{3}\,em\,\boxed{1}\,temos:\\\sf c^x=\dfrac{c^y}{c^z}\\\\\sf c^x=c^{y-z}\implies x=y-z~\blacksquare\end{array}}$

$\large\boxed{\begin{array}{l}\sf 3)\log_ca^n=n\cdot log_ca.\\\underline{\sf demonstrac_{\!\!,}\tilde ao\!:}\\\sf fac_{\!\!,}a~ \log_ca^n=x~e~\log_ca=y\\\sf vamos\,provar\,que\,x=n\cdot y.\\\sf Com\,efeito, \log_ca^n=x\implies a^n=c^x.\\\sf \log_ca=y\implies a=c^y\\\sf a^n=c^x\\\sf (c^y)^n=c^x\\\sf c^{n\cdot y}=c^x\implies x=n\cdot y~\blacksquare\end{array}}$$\large\boxed{\begin{array}{l}\tt a)~\sf S=\log_44-\log_66^2+3^{\log_34}\\\rm utilizando\,as\,propriedades\,j\acute a\,demonstradas\\\rm anteriormente\,temos\!:\\\sf S=\log_44-2\log_66+3^{\log_34}\\\sf S=1-2\cdot1+4\\\sf S=1-2+4\\\sf S=3\end{array}}$

$\large\boxed{\begin{array}{l}\tt b)~\sf S=\log_{\frac{1}{3}}\bigg(\dfrac{1}{3}\bigg)^5+\log_71-25^{\log_52}\\\underline{\bf C\acute alculos\,auxiliares:}\\\\\sf S=\log_{\frac{1}{3}}\bigg(\dfrac{1}{3}\bigg)^5=5\log_{\frac{1}{3}}\dfrac{1}{3}=5\cdot1=5\\\\\sf \log_71=0\\\sf 25^{\log_52}=(5^2)^{\log_52}=5^{2\log_52}\\\sf 25^{\log_52}=(5^{\log_52})^2=2^2=4\\\sf vamos\,substituir:\\\sf S=\log_{\frac{1}{3}}\bigg(\dfrac{1}{3}\bigg)^5-\log_71-25^{\log_52}\\\\\sf S=5-0-4\\\sf S=1\end{array}}$

$\large\boxed{\begin{array}{l}\rm 7)~~Qual\,\acute e\,o\,valor\,num\acute erico\,de\,\log_{\dfrac{1}{3}}(4^{\log_29})?\\\underline{\bf soluc_{\!\!,}\tilde ao\!:}\\\sf 4^{\log_29}=(2^2)^{\log_29}=(2^{\log_29})^2=9^2=81.\\\sf\log_{\dfrac{1}{3}}(4^{\log_29})=\log_{\dfrac{1}{3}}81=\log_{\dfrac{1}{3}}\bigg(\dfrac{1}{3}\bigg)^{-4}\\\\\sf \log_{\dfrac{1}{3}}(4^{\log_29})=-4\log_{\dfrac{1}{3}}\dfrac{1}{3}=-4\cdot1=-4\end{array}}$