Question

log in baza 2 din x + log in baza 4 din x + log in baza 8 din x = 11/6

Answer 1

[tex]\displaystyle \log_2{x}+\log_4{x}+\log_8{x}=\frac {11}6, x\ \textgreater \ 0\\ \\ \log_2{x}+\log_{2^2}{x}+\log_{2^3}{x}=\frac {11}6\\ \\ \\ Folosind~formula:~\log_{a^x}{n}=\frac 1x\log_{a}{n}\\ \\ \\ \log_2{x}+\frac 12\log_2{x}+\frac 13\log_2{x}=\frac {11}6\ \\ \frac{11}6\log_2{x}=\frac {11}6\\ \\ \log_2{x}=1\\ \\ x=2.\\ \\ S={2}[/tex]

Answer 2

$\log_{\big{a^n}}b = \dfrac{1}{\log_{\big{b}}a^n} = \dfrac{1}{n\cdot \log_{\big{b}}a} = \dfrac{1}{n}\cdot \dfrac{1}{\log_{\big{b}}a} = \dfrac{1}{n}\cdot \log_{\big{a}}b \\ \\ \Rightarrow \boxed{\log_{\big{a^n}}b = \dfrac{1}{n}\cdot \log_{\big{a}}b}\rightarrow (o sa ne folosim de formula asta)$


$\\\log_{\big2}x+\log_{\big4}x+\log_{\big8}x=\dfrac{11}{6} \\ \\ \log_{\big2}x+\log_{\big{2^2}}x+\log_{\big{2^3}}x=\dfrac{11}{6}\\ \\ \log_{\big2}x+\dfrac{1}{2}\cdot \log_{\big2}x +\dfrac{1}{3}\cdot \log_{\big2}x = \dfrac{11}{6}\Big|\cdot 6 \\ \\ 6\cdot\log_{\big2}x+3\cdot \log_{\big2}x+2\cdot \log_{\big2}x = 11 \\ \\ \log_{\big2}x \cdot(6+3+2) = 11 \\ \\ (\log_{\big2}x)\cdot 11=11 \\ \\ \log_{\big2}x = \dfrac{11}{11} \\ \\ \log_{\big2}x = 1 \\ \\ x = 2^1 \\ \\ x = 2$