Question

1) dados log2 = 0,30103 e log3 = 0,47712 então o log 7,2 é?

2) sabendo que log a = L e que log b = M, então o log de a na base b é?

3) a solução da equação log x^2+log x=1 é?

Answer 1

1)

$7,2=\frac{2*2*2*3*3}{10}= \frac{2^3.3^2}{10} \\\\\\log \;7,2=log\;\frac{2^3.3^2}{10}\\\\\left(log\;2^3+log\;3^2\right)-log\;10\\\\\\\left(3log\;2+2log\;3\right)-log\;10\\\\\left(3*0,30103+2*0,47712\right)-1\\\\\\1,85733-1\\\\\\0,85733$

2) Basta mudar a base

$log_b\;a=\frac{log_{10}\;a}{log_{10}\;b} \\\\log_b\;a=\frac{L}{M}$

3)

$log\;x^2+log\;x=1\\\\2log\;x+log\;x=1\\\\3log\;x=1\\\\log\;x=\frac{1}{3}\\ \\x = 10^{\frac{1}{3}}\\\\x=\sqrt[3]{10}$