Matemática, perguntado por clarabeyknowles, 1 ano atrás

O valor de log3 27 + log5 625 + log7 49 + log10 0,001 + log6 1/6 é:

Soluções para a tarefa

Respondido por Lukyo
1
E=\mathrm{\ell og}_{3\,}27+\mathrm{\ell og}_{5\,}625+\mathrm{\ell og}_{7\,}49+\mathrm{\ell og}_{10\,}0,001+\mathrm{\ell og}_{6\,}\dfrac{1}{6}\\ \\ \\ E=\mathrm{\ell og}_{3\,}(3^{3})+\mathrm{\ell og}_{5\,}(5^{4})+\mathrm{\ell og}_{7\,}(7^{2})+\mathrm{\ell og}_{10\,}(10^{-4})+\mathrm{\ell og}_{6\,}(6^{-1})\\ \\ E=3\mathrm{\,\ell og}_{3\,}3+4\mathrm{\,\ell og}_{5\,}5+2\mathrm{\,\ell og}_{7\,}7-4\mathrm{\,\ell og}_{10\,}10-1\mathrm{\,\ell og}_{6\,}6\\ \\ E=3\cdot 1+4\cdot 1+2\cdot 1-4\cdot 1-1\cdot 1\\ \\ E=3+4+2-4-1\\ \\ E=4

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