Question

$se \: log(8) = a \: entao \: log(5) \: vale$
$a) {a}^{3} \\ b)5a - 1 \\ c) \frac{2a}{3} \\ d) \frac{1 + a}{3} \\ e) \frac{1 - a}{3}$
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Answer 1

Resposta:

Explicação passo-a-passo:

Resposta:

Explicação passo-a-passo:

$log_{b}(x\cdot y)=log_{b}(x)+log_{b}(y)\\\\\\log_{b}(a)=\dfrac{log_{c}(a)}{log_{c}(b)}~~~(mudanca~de~base~de~'b'~para~'c')\\\\\\log_{b}(a)=\dfrac{1}{log_{a}(b)}$

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$log_{10}(8)=a\\\\\\\dfrac{1}{log_{8}(10)}=a\\\\\\\dfrac{1}{log_{8}(2\cdot5)}=a\\\\\\\dfrac{1}{log_{8}(2)+log_{8}(5)}=a\\\\\\\dfrac{1}{(\frac{1}{3})+log_{8}{5}}=a$

Vamos multiplicar em cruz:

$a\cdot\left[\dfrac{1}{3}+log_{8}(5)\right]=1\\\\\\\dfrac{1}{3}+log_{8}(5)=\dfrac{1}{a}\\\\\\log_{8}(5)=\dfrac{1}{a}-\dfrac{1}{3}\\\\\\log_{8}(5)=\dfrac{3-a}{3a}$

Vamos mudar a base para 10:

$\dfrac{log~5}{log~8}=\dfrac{3-a}{3a}$

Como log 8 = a:

$\dfrac{log~5}{a}=\dfrac{3-a}{3a}\\\\\\\boxed{\boxed{log~5=\dfrac{3-a}{3}}}$