Question

São dados: $log_{15} 3 = a$ e $log_{15} 2 = b$. O valor de log 2 é:
a)$\frac{a}{1-a+b}$ d)$\frac{a}{1+a-b}$

b)$\frac{b}{1-a+b}$ e)$\frac{b}{a-b-1}$

c)$\frac{b}{1+a-b}$

Answer 1

Ola Vinicius

a = log15(3)
b = log15(2) 

10 = 15*2/3 

log15(10) = log15(15) - log15(3) + log15(2)
log15(10) = 1 + b - a

log10(2) = log15/2)/log15(10) = b/(1 + a + b) (B)