Question

Incógnita em um logaritmo. Como se resolve isso?

Answer 1

$2,6=\dfrac{2}{3}\mathrm{\ell og_{10}}\left(\dfrac{E}{7\times 10^{-3}} \right )\\ \\ \dfrac{2,6 \cdot 3}{2}=\mathrm{\ell og_{10}}\left(E \right )-\mathrm{\ell og_{10}}\left(7\times 10^{-3} \right )\\ \\ 3,9=\mathrm{\ell og_{10}}\left(E \right )-\mathrm{\ell og_{10}}\left(7\right)-\mathrm{\ell og_{10}} \left(10^{-3} \right )\\ \\ 3,9=\mathrm{\ell og_{10}}\left(\dfrac{E}{7} \right )+3\,\mathrm{\ell og_{10}} \left(10 \right )\\ \\ 3,9=\mathrm{\ell og_{10}}\left(\dfrac{E}{7} \right )-3\\ \\ \mathrm{\ell og_{10}}\left(\dfrac{E}{7} \right )=3,9-3\\ \\ \mathrm{\ell og_{10}}\left(\dfrac{E}{7} \right )=0,9\\ \\ \dfrac{E}{7}=10^{0,9}\\ \\ \boxed{E=7\cdot 10^{0,9} \approx 55,6}$