Witam, proszę o szybkie rozwiązanie rownań logarytmicznych:.... Question from @westen

westen @westen

November 2018 1 23 Report

Witam, proszę o szybkie rozwiązanie rownań logarytmicznych:

$b) (lnx)(1+lnx)=2+lnx^2$

$c) 2log3 - \frac{1}{2}log_{0,1}x=log_{100}0,5$

$d) log_{0,2}49 + 2log_{25}x = 3log_{125}7$

$e) 1-lnx=3ln2$

kaos $lnx + (lnx)^2=2+2lnx\\ lnx=y \\y+y^2=2+2y\\y^2-y-2=0\\y_{1}=-1; y_{2}=2\\x=e^{y}\\x_{1}=e^{-1};x_{2}=e^2 \\\\ c) 2log3- \frac{1}{2}\frac{log_{10}{x}}{log_{10}{0,1}}=\frac{log_{10}{0,5}}{log_{10}{100}}\\ 2log3-\frac{1}{2} \frac{log_{10}{x}}{-1}=\frac{-log2}{log2}\\2log3+log(\sqrt{x})=-1 \\ 2log3+log(\sqrt{x})=log10^{-1} \\ log(9\sqrt{x})=log(10^{-1}) \\9\sqrt{x}=10^{-1} \\ \sqrt{x} =\frac{1}{90} \\ x=90^{-2}$

$d) log_{0,2}{49}=log_{5^{-1}}{49}=\frac{log_{7}{49}}{log_{7}{5^{-1}}}=\frac{2}{-log_{7}{5}}=-2log_{5}{7}\\ log_{125}{7}=\frac{log_{5}{7}}{log_{5}{125}}=\frac{log_{5}{7}}{3}\\ log_{25}{x}=\frac{log_{5}{x}}{log_{5}{25}}=\frac{log_{5}{x}}{2} \\ -2log_{5}{7}+2*\frac{log_{5}{x}}{2}=3*\frac{log_{5}{7}}{3}\\log_{5}{x}=3log_{5}{7}\\x=7^3$

$e) 1=ln(e); ln(e)-ln(x)=ln(2^3)\\\frac{e}{x}=8 \\ x=\frac{e}{8}$

2 votes Thanks 2