Question

provr that log[x+√x2-1]+log[x-√x2-1]=0​

Answer 1

Resposta:

Explicação passo-a-passo:

loga + logb= logab

(a + b)(a - b) = a² - b²

log[x + √(x²- 1) + log[x - √(x² - 1) = log[x + √(x²- 1)][x - √(x² - 1)] =

= log[x²- (√x² - 1)]² = log[x² - (x² - 1)] = logx² - x² + 1 = log1 = 0

how we wanted to prove