Se a^-1 +b^-1 = 2 e 5^a =125^b = ?

 

Se a^-1 +b^-1 = 2  

Encontre: 5^a =125^b = ?

1/a + 1/b = 2

5^a = 125^b = ?

5^a = 125^b = x

5^a = x

log 5^a = log x

a log 5 = log x

a = log x/log 5

a = log 5  x

125^b = x

log 125^b = log x

b log 125 = log x

b = log x / log 125

b = log 125 x

========== 

1/a + 1/b = 2

1/ (log 5 x) + 1/(log 125 x) = 2

log x 5  + log x 125 = 2

log x (5 *125) = 2

log x 625 = 2

x² = 625

 x = 25