Question

помогите решить пожалуйста

Answer 1

1. log₆216 · log₉729 = log₆6³ · log₉9³ = 3·3 = 9

$2.~13cdot 10^{log_{10}2}=13cdot 2 = 26\\3.~9^{log_37}=(3^2)^{log_37}=(3^{log_37})^2=7^2=49\\4.~log_{0,2}125=log_{frac 15}125=log_{5^{-1}}125=-log_55^3=-3\\5.~log_{25}0,2=log_{5^2}dfrac 15=dfrac 12log_55^{-1}=-dfrac 12\\6.~log_6234-log_66,5=log_6dfrac {234}{6,5}=log_636=log_66^2=2\\7.~log_{25}25+log_{0,2}625=1+log_{frac 15}625=1-log_55^4=1-4=-3\\8.~log_{0,3}10-log_{0,3}3=log_{0,3}dfrac {10}3=log_{0,3}(0,3)^{-1}=-1$

$9.~dfrac{log_64}{log_62}=log_24=log_22^2=2\\10.~dfrac{log_92}{log_{81}2}=dfrac{log_92}{log_{9^2}2}=dfrac{log_92}{frac 12log_92}=2\\11.~log_413cdot log_{13}16=dfrac 1{log_{13}4}cdot log_{13}4^2=dfrac {2log_{13}4}{log_{13}4}=2\\12.~dfrac {2^{log_{13}507}}{2^{log_{13}3}}=2^{log_{13}507-log_{13}3}=2^{log_{13}(507:3)}=2^{log_{13}169}=\\~~~=2^{log_{13}13^2}=2^2=4$

$13.~(1-log_848)(1-log_648)=(1-log_8(8cdot6))(1-log_6(6cdot8))=\\~~~=(1-log_88-log_86))(1-log_66-log_68))=\\~~~=(1-1-log_86)(1-1-log_68)=(-log_86)(-log_68)=\\~~~=log_86cdotlog_68=dfrac{log_86}{log_86}=1\\14.~50log_{10}sqrt[5]10=50log_{10}10^{frac 15}=50cdot dfrac 15=10\\15.~log_{sqrt[5]10}10=log_{10^frac 15}10=dfrac 1{frac 15}log_{10}10=5$

$16.~dfrac {log_252}{2+log_213}=dfrac {log_252}{log_24+log_213}=dfrac {log_252}{log_2(4cdot13)}=dfrac {log_252}{log_252}=1$