Question

considere f(x) log2 (x² + 3x) calcule f(5) sendo que log 2 =0,301

Answer 1

Se $f(x)=\mathrm{\ell og}_{2\,}(x^{2}+3x),$ então

$f(5)=\mathrm{\ell og}_{2\,}(5^{2}+3\cdot 5)\\ \\ f(5)=\mathrm{\ell og}_{2\,}(25+15)\\ \\ f(5)=\mathrm{\ell og}_{2\,}(40)\\ \\ f(5)=\mathrm{\ell og}_{2\,}(2^{2}\cdot 10)\\ \\ f(5)=\mathrm{\ell og}_{2\,}(2^{2})+\mathrm{\ell og}_{2}\,10\\ \\ f(5)=2\,\mathrm{\ell og}_{2\,}2+\mathrm{\ell og}_{2}\,10\\ \\ f(5)=2\cdot 1+\mathrm{\ell og}_{2}\,10\\ \\ f(5)=2+\mathrm{\ell og}_{2}\,10$


Pela lei de mudança de base, temos que

$\mathrm{\ell og}_{2}\,10=\dfrac{\mathrm{\ell og\,}10}{\mathrm{\ell og\,}2}\\ \\ \\ \mathrm{\ell og}_{2}\,10=\dfrac{1}{\mathrm{\ell og\,}2}$


Sendo assim, temos que

$f(5)=2+\dfrac{1}{\mathrm{\ell og\,}2}$


Substituindo o valor dado, chegamos a

$f(5)=2+\dfrac{1}{0,301}\\ \\ \\ f(5)\approx 5,32$