Question

Calcule a soma log2 16+log3 81+log4 0,25?

Answer 1

$observe que: 0,25= \frac{25}{100} = \frac{1}{4}= 4^{-1}\\ \log_2 \ 16 \ + \ \log_3 \ 81 \ + \log_4 \ 0,25\\ = \log_2 \ 2^4 \ + \ \log_3 \ 3^4 \ + \log_4 \ 4^{-1}\\ = 4\cdot\log_2 \ 2 \ + 4\cdot\ log_3 \ 3 \ + (-1)\cdot \log_4 \ 4\\ = 4\cdot1\ + 4\cdot\ 1\ + (-1)\cdot \ 1\\ = 4 + 4 -1\\ =7$