Question

Problema 6.5 Sabendo-se que loga(x) = 2, logb(x) = 3 e logc(x) = 5, calcule
(a) logab (x)
(b) logabc (x)
(c) logabc (x)

Answer 1

Propriedade usada (vem da mudança de base)

$\boxed{\boxed{log_{b}(a)=\dfrac{1}{log_{a}(b)}}}$
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$log_{a}(x)=2~~~\therefore~~~log_{x}(a)=\dfrac{1}{2}\\\\\\log_{b}(x)=3~~~\therefore~~~log_{x}(b)=\dfrac{1}{3}\\\\\\log_{c}(x)=5~~~\therefore~~~log_{x}(c)=\dfrac{1}{5}$
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Letra A:

$log_{ab}(x)=\dfrac{1}{log_{x}(ab)}\\\\\\log_{ab}(x)=\dfrac{1}{log_{x}(a)+log_{x}(b)}\\\\\\log_{ab}(x)=\dfrac{1}{(\frac{1}{2})+(\frac{1}{3})}\\\\\\log_{ab}(x)=\dfrac{1}{(\frac{3+2}{3\cdot2})}\\\\\\log_{ab}(x)=\dfrac{1}{(\frac{5}{6})}\\\\\\\boxed{\boxed{log_{ab}(x)=\dfrac{6}{5}}}$
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Letra B:

$log_{abc}(x)=\dfrac{1}{log_{x}(abc)}\\\\\\log_{abc}(x)=\dfrac{1}{log_{x}(a)+log_{x}(b)+log_{x}(c)}\\\\\\log_{abc}(x)=\dfrac{1}{(\frac{1}{2})+(\frac{1}{3})+(\frac{1}{5})}\\\\\\log_{abc}(x)=\dfrac{1}{(\frac{15}{30})+(\frac{10}{30})+(\frac{6}{30})}\\\\\\log_{abc}(x)=\dfrac{1}{(\frac{31}{30})}\\\\\\\boxed{\boxed{log_{abc}(x)=\dfrac{30}{31}}}$
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Letra C:

$log_{bc}(x)=\dfrac{1}{log_{x}(bc)}\\\\\\log_{bc}(x)=\dfrac{1}{log_{x}(b)+log_{x}(c)}\\\\\\log_{bc}(x)=\dfrac{1}{(\frac{1}{3})+(\frac{1}{5})}\\\\\\log_{bc}(x)=\dfrac{1}{(\frac{8}{15})}\\\\\\\boxed{\boxed{log_{bc}(x)=\dfrac{15}{8}}}$

ou:

$log_{ac}(x)=\dfrac{1}{log_{x}(ac)}\\\\\\log_{ac}(x)=\dfrac{1}{log_{x}(a)+log_{x}(c)}\\\\\\log_{ac}(x)=\dfrac{1}{(\frac{1}{2})+(\frac{1}{5})}\\\\\\log_{ac}(x)=\dfrac{1}{(\frac{7}{10})}\\\\\\\boxed{\boxed{log_{ac}(x)=\dfrac{10}{7}}}$