log-2-x-log-4-x-log-16-x-7-

Question Number 32960 by mondodotto@gmail.com last updated on 08/Apr/18

\log_{2} x + \log_{4} x + \log_{16} x = 7

Answered by Joel578 last updated on 08/Apr/18

\log_{a^{n}} b^{m} = \frac{m}{n} \log_{a} b

\log_{2} x + \log_{2^{2}} x + \log_{2^{4}} x = 7

\log_{2} x + \frac{1}{2} \log_{2} x + \frac{1}{4} \log_{2} x = 7

\frac{7}{4} \log_{2} x = 7

\log_{2} x = 4

x = 2^{4} = 16

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