Sendo logb a=4 e logb c=1, encontre o valor de:
![logb \: (\sqrt{b} \times c) logb \: (\sqrt{b} \times c)](https://tex.z-dn.net/?f=logb+%5C%3A+%28%5Csqrt%7Bb%7D++%5Ctimes+c%29)
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Essas aqui são as propriedades que eu vou usar:
![\log_c{(a\cdot b)}=\log_ca+\log_{c}b \log_c{(a\cdot b)}=\log_ca+\log_{c}b](https://tex.z-dn.net/?f=%5Clog_c%7B%28a%5Ccdot+b%29%7D%3D%5Clog_ca%2B%5Clog_%7Bc%7Db)
![\log_{c}{a^b}=b\cdot\log_{c}{a} \log_{c}{a^b}=b\cdot\log_{c}{a}](https://tex.z-dn.net/?f=%5Clog_%7Bc%7D%7Ba%5Eb%7D%3Db%5Ccdot%5Clog_%7Bc%7D%7Ba%7D)
![\displaystyle \log_{a}{a}=1 \displaystyle \log_{a}{a}=1](https://tex.z-dn.net/?f=+%5Cdisplaystyle+%5Clog_%7Ba%7D%7Ba%7D%3D1+)
![\displaystyle \sqrt[m]{a^n}=a^{\frac{n}{m}} \displaystyle \sqrt[m]{a^n}=a^{\frac{n}{m}}](https://tex.z-dn.net/?f=+%5Cdisplaystyle+%5Csqrt%5Bm%5D%7Ba%5En%7D%3Da%5E%7B%5Cfrac%7Bn%7D%7Bm%7D%7D+)
Com elas, dá para resolver,
![\displaystyle \log_{b}{(\sqrt{b}\cdot c)} \displaystyle \log_{b}{(\sqrt{b}\cdot c)}](https://tex.z-dn.net/?f=+%5Cdisplaystyle+%5Clog_%7Bb%7D%7B%28%5Csqrt%7Bb%7D%5Ccdot+c%29%7D+)
![\displaystyle \log_{b}{\sqrt{b}}+\log_{b}{c} \displaystyle \log_{b}{\sqrt{b}}+\log_{b}{c}](https://tex.z-dn.net/?f=+%5Cdisplaystyle+%5Clog_%7Bb%7D%7B%5Csqrt%7Bb%7D%7D%2B%5Clog_%7Bb%7D%7Bc%7D+)
![\displaystyle \log_{b}{b^{\frac{1}{2}}}+1 \displaystyle \log_{b}{b^{\frac{1}{2}}}+1](https://tex.z-dn.net/?f=+%5Cdisplaystyle+%5Clog_%7Bb%7D%7Bb%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%2B1+)
![\displaystyle \frac{1}{2} \cdot \log_{b}{b}+1 \displaystyle \frac{1}{2} \cdot \log_{b}{b}+1](https://tex.z-dn.net/?f=+%5Cdisplaystyle+%5Cfrac%7B1%7D%7B2%7D+%5Ccdot+%5Clog_%7Bb%7D%7Bb%7D%2B1+)
![\displaystyle \frac{1}{2} \cdot 1+1 \displaystyle \frac{1}{2} \cdot 1+1](https://tex.z-dn.net/?f=+%5Cdisplaystyle+%5Cfrac%7B1%7D%7B2%7D+%5Ccdot+1%2B1+)
![\displaystyle \frac{1+2}{2} \displaystyle \frac{1+2}{2}](https://tex.z-dn.net/?f=+%5Cdisplaystyle+%5Cfrac%7B1%2B2%7D%7B2%7D+)
Com elas, dá para resolver,
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