Question

Log in baza 2 din x+ log in baza 4 din x+ log in baza 8 din x =5,5

Answer 1

 

$\displaystyle\bf\\log_{2}(x)+log_{4}(x)+log_{8}(x)=5,\!5\\\\log_{2}(x)+log_{2^2}(x)+log_{2^3}(x)=5,\!5\\\\log_{2}(x)+\frac{1}{2}log_{2}(x)+\frac{1}{3}log_{2}(x)=5,\!5\\\\\left(1+\frac{1}{2}+\frac{1}{3}\right)\cdot log_{2}(x)=5,\!5\\\\\left(\frac{6}{6}+\frac{3}{6}+\frac{2}{6}\right)\cdot log_{2}(x)=5,\!5\\\\\frac{6+3+2}{6}\cdot log_{2}(x)=5,\!5\\\\\frac{11}{6}\cdot log_{2}(x)=5,\!5\\\\log_{2}(x)=\frac{~~5,\!5~~}{\dfrac{11}{6}}\\\\log_{2}(x)=\frac{5,\!5\cdot6}{11}=\frac{33}{11}=3\\\\\implies x=8$