Question

logaritmo: 16^(1+log 5 na base 2)?

Answer 1

$16^{1+\mathrm{\ell og}_{2\,}5}\\ \\ =16^{1}\cdot 16^{\mathrm{\ell og}_{2\,}5}\\ \\ =16\cdot \left(2^{4}\right)^{\mathrm{\ell og}_{2\,}5}\\ \\ =16\cdot 2^{4\mathrm{\,\ell og}_{2\,}5}\\ \\ =16\cdot \left(2^{\mathrm{\ell og}_{2\,}5} \right )^{4}$


Se fizermos $x=\mathrm{\ell og}_{2\,}5$, então

$2^{x}=5\\ \\ 2^{\mathrm{\ell og}_{2\,}5}=5$


Voltando à expressão, temos

$16\cdot \left(2^{\mathrm{\ell og}_{2\,}5} \right )^{4}\\ \\ =16 \cdot 5^{4}\\ \\ =2^{4} \cdot 5^{4}\\ \\ =\left(2 \cdot 5 \right )^{4}\\ \\ =10^{4}\\ \\ =10\,000\\ \\ \\ \boxed{16^{1+\mathrm{\ell og}_{2\,}5}=10\,000}$