Question

Se log8 = a entao log5 vale?

Answer 1

Propriedades usadas:

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

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

Answer 2

 log8 = a    log5 ?sabemos quando a base tem nada vale 10
 8 = 2³

log10 ^2^3  = a  (pela propriedade podemos chutar o 3 pra frente.)
 3log10² = a       ( 3 passa dividindo )

log10² = a/3

5 = 10/2
                     propriedade da divisão
log^5 = logo10^10/2 =   log10^10 - loga10² =  1 - a/3   mmc =3

log^5 = 3 - a
               3