Question

sendo loga x=2 logb x=5 e logc x=6 calcular logabcx

Answer 1

Explicação passo-a-passo:

Temos:

$log_{a}(x) = 2$

$log_{b}(x) = 5$

$log_{c}(x) = 6$

Tomando

$log_{abc}(x)$

Vamos mudar a base para x:

$log_{abc}(x) = \frac{ log_{x}(x) }{ log_{x}(abc) }$

$log_{abc}(x) = \frac{1}{ log_{x}(abc) }$

$log_{abc}(x) = \frac{1}{ log_{x}(a) + log_{x}(b) log_{x}(c) }$

$log_{abc}(x) = \frac{1}{2 + 5 + 6}$

$log_{abc}(x) = \frac{1}{13}$