Question

se log2 = 0,3 e log3= 0,48 e log7= 0,84, o valor de x para que 15^x = 42, é

Answer 1

Olá Jonathan,


Organizando as informações:

$\mathsf{\ell og(2)\approx 0,3}\\\mathsf{\ell og(3)\approx 0,48}\\\mathsf{\ell og(7)\approx 0,84}\\\\\\\mathsf{15^x=42\Rightarrow \ell og(15^x)=\ell og(42)\Rightarrow \ell og\Big(\Big[\dfrac{3\cdot 10}{2}\Big]^x\Big)=\ell og(3\cdot2\cdot7)}\\\\=\\\\\mathsf{x\cdot[\ell og(3)+\ell og(10)-\ell og(2)]=\ell og(3)+\ell og(2)+\ell og(7)}\\\\=\\\\\mathsf{x\cdot[0,48+1-0,3]=0,48+0,3+0,84\Rightarrow 1,18x=1,62}\\\\=\\\\\mathsf{x=\dfrac{1,62}{1,18}\Rightarrow \boxed{\mathsf{x\approx 1,37}}}$

Dúvidas? comente

Answer 2

log2 = 0,3 e log3= 0,48 e log7= 0,84, o valor de x para que 15^x = 42, é

log 42 = x.      Mudança de base 10.
     15

x = log 42  ==> 1,62  ≈1,37
      log 15          1,18

log42 = log2 + log3 + log7 ==> 0,3 + 0,48 + 0,84 ==.> 1,62

log 15 = log 3 - log5 ==>  x = log3 + log10 - log2 ==> 0,48 + 1 - 0,3 ==> 1,18