Question

sabendo-se que: logx a = 6, logx b = 3 e log x c = 2 calcular:

a) log x a⁴/ b³. c²
b) logx 3√ac/ b​

Answer 1

Resposta:

Explicação passo a passo:

a)

logₓa⁴/b³c²=logₓa⁴-logₓb³c²=4logₓa-(logₓb³+logₓc²)=4logₓa-(3logₓb+2logₓc)=4.6-(3.3+2.2)=24-(9+4)=24-13=11

b)

logₓ∛ac/b=logₓ(ac/b)¹/³=logₓ(ac/b)/3=[logₓ(ac)-logₓb]/3=[logₓa+logₓc-logₓb]/3=[6+2-3]/3=5/3

Propriedades:

logₓ(a)ⁿ=nlogₓ(a)

logₓ(a.b)=logₓ(a)+logₓ(b)

logₓ(a/b)=logₓ(a)-logₓ(b)