log-a-x-8-log-b-x-3-log-c-x-6-log-abc-x-

Question Number 120120 by benjo_mathlover last updated on 29/Oct/20

{\begin{cases} \log_{a} (x) = 8 \\ \log_{b} (x) = 3 \\ \log_{c} (x) = 6 \end{cases} \Rightarrow \log_{a b c} (x) = ?

Answered by bemath last updated on 29/Oct/20

{\begin{cases} \log_{a} (x) = 8 \Rightarrow \log_{x} (a) = \frac{1}{8} \\ \log_{b} (x) = 3 \Rightarrow \log_{x} (b) = \frac{1}{3} \\ \log_{c} (x) = 6 \Rightarrow \log_{x} (c) = \frac{1}{6} \end{cases}

T h e n \log_{a b c} (x) = \frac{1}{\log_{x} (a b c)} = \frac{1}{\frac{1}{8} + \frac{1}{3} + \frac{1}{6}}

= \frac{24}{3 + 8 + 4} = \frac{24}{15} = \frac{8}{5}

Answered by $@y@m last updated on 29/Oct/20

x = a^{8} = b^{3} = c^{6}

a = x^{\frac{1}{8}}, b = x^{\frac{1}{3}}, c = x^{\frac{1}{6}}

a b c = x^{\frac{1}{8} + \frac{1}{3} + \frac{1}{6}} = x^{\frac{15}{24}}

\log_{x} (a b c) = \frac{15}{24}

\log_{a b c} (x) = \frac{24}{15} = \frac{8}{5}

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