Question

1) log7 49+ log2 128 é?

2) o log de 8 na base 1/2 é?

Answer 1

1)

$log_7\,49=x\\\\49=7^x\\\\7^2=7^x\\\\x=2\\\\\\log_2\,128=y\\\\128=2^y\\\\2^7=2^y\\\\y=7$

Assim:

log7 49+ log2 128 = 2 + 7 = 9

2)

$log_{\frac{1}{2}}8=x\\\\8=\left(\frac{1}{2}\right)^x\\\\2^3=\left(\frac{1}{2}\right)^x\\\\2^3=\left(\frac{2}{1}\right)^{-1^x}\\\\2^3=2^{-1*x}\\\\2^3=2^{-x}\\\\-x=3\\\\x=-3$