Question

Sabendo - se que logx a =8, logx b=2 e logx c=1, calcular:logx a3/b2 x c4

Answer 1

P(z) (x-1) + d = p(x) = ax³+bx²+cx+d
P(z) (x-1) +d = ax³+bx²+cx+d
P(z) (x-1) = ax³+bx²+cx+d-d
P(z) (x-1) = ax³+bx²+cx
para x = 1 =>P(z) (x-1)= 0 = a(1)³+b(1)²+c(1)
para x = 1 => 0 = a + b + c

Answer 2

Explicação passo-a-passo:

Lembre-se que:

• $\sf log_{b}~(a\cdot c)=log_{b}~a+log_{b}~c$

• $\sf log_{b}~(a\div c)=log_{b}~a-log_{b}~c$

• $\sf log_{b}~a^m=m\cdot log_{b}~a$

Assim:

$\sf log_{x}~\Big(\dfrac{a^3}{b^2\cdot c^4}\Big)=log_{x}~a^3-log_{x}~(b^2\cdot c^4)$

$\sf log_{x}~\Big(\dfrac{a^3}{b^2\cdot c^4}\Big)=log_{x}~a^3-(log_{x}~b^2+log_{x}~c^4)$

$\sf log_{x}~\Big(\dfrac{a^3}{b^2\cdot c^4}\Big)=log_{x}~a^3-log_{x}~b^2-log_{x}~c^4$

$\sf log_{x}~\Big(\dfrac{a^3}{b^2\cdot c^4}\Big)=3\cdot log_{x}~a-2\cdot log_{x}~b-4\cdot log_{x}~c$

$\sf log_{x}~\Big(\dfrac{a^3}{b^2\cdot c^4}\Big)=3\cdot8-2\cdot2-4\cdot1$

$\sf log_{x}~\Big(\dfrac{a^3}{b^2\cdot c^4}\Big)=24-4-4$

$\sf \red{log_{x}~\Big(\dfrac{a^3}{b^2\cdot c^4}\Big)=16}$