Question

Sabendo que 10 p = 3 e que 10q = 2 , calcule em função de p e q o valor do log

Answer 1

$\displaystyle \sf 10^P =3 \to \log 10^P = \log 3 \to \boxed{\sf P = \log 3 } \\\\ 10^q=2 \to \log 10^q=\log 2\to \boxed{\sf q = \log 2} \\\\ temos : \\\\ \log_{\sqrt[4]{0,015}} (225000000) \\\\\\ \log_{\displaystyle (15\cdot 10^{-3})^{\frac{1}{4}}}(15\cdot 10^3)^2 \\\\\\ \frac{4}{1} \cdot 2\cdot \log_{(15\cdot 10^{-3})}(15\cdot 10^3 ) \\\\\\ 8\cdot \frac{\log (15\cdot 10^3 )}{\log (15\cdot 10^{-3})} \\\\\\ 8\cdot \left[\frac{\log(15)+\log (10)^3}{\log(15)+\log(10)^{-3}} \right]$

$\displaystyle \sf 8\cdot \left[\frac{\displaystyle \log\left(\frac{30}{2} \right)+3\cdot \log 10 }{\displaystyle \log\left(\frac{30}{2} \right)-3\cdot \log 10 } \right] \\\\\\ 8\cdot \left[\frac{\log 30-\log 2 +3 }{\log 30-\log 2 -3 }\right] \\\\\\ 8\cdot \left[\frac{\log (3\cdot 10)-q+3 }{\log (3\cdot 10)-q-3}\right] \\\\\\ 8\cdot\left[ \frac{\log 3 +\log 10-q+3}{\log 3+\log 10-q-3} \right] \\\\\\ 8\cdot \left [\frac{p+1-q+3}{p+1-q-3} \right] \\\\\\ \frac{8\cdot (p-q+4)}{p-q-2} }$

Portanto :

$\boxed{\sf \displaystyle \sf \log_{\sqrt[4]{0,015}}(225000000) = \frac{8(p-q+4)}{p-q-2} \ }\checkmark$