Question

Log din baza x din 2 + Log din baza radical x din 2=9

Answer 1

Aceasta este rezolvarea.

Answer 2

$\log _x\left(2\right)+\log _{\sqrt{x}}\left(2\right)=9 \\ \\ \log _{\sqrt{x}}\left(2\right)=\frac{\ln \left(2\right)}{\ln \left(\sqrt{x}\right)} \\ \\ \log _x\left(2\right)+\frac{\ln \left(2\right)}{\ln \left(\sqrt{x}\right)}=9 \\ \\ \log _x\left(2\right)=\frac{\ln \left(2\right)}{\ln \left(x\right)} \\ \\ \frac{\ln \left(2\right)}{\ln \left(x\right)}+\frac{\ln \left(2\right)}{\ln \left(\sqrt{x}\right)}=9 \\ \\ \ln \left(\sqrt{x}\right)=\frac{1}{2}\ln \left(x\right)$$\frac{\ln \left(2\right)}{\ln \left(x\right)}+\frac{\ln \left(2\right)}{\frac{1}{2}\ln \left(x\right)}=9 \\ \\ \ln \left(x\right)=u \\ \\ \frac{\ln \left(2\right)}{u}+\frac{\ln \left(2\right)}{\frac{1}{2}u}=9 \\ \\ \frac{\ln \left(2\right)}{u}\cdot \frac{u^2}{2}+\frac{\ln \left(2\right)}{\frac{1}{2}u}\cdot \frac{u^2}{2}=9\cdot \frac{u^2}{2} \\ \\ \frac{\ln \left(2\right)u}{2}+\ln \left(2\right)u=\frac{9u^2}{2} \\ \\ \frac{\ln \left(2\right)u}{2}+\ln \left(2\right)u=\frac{9u^2}{2}$$\frac{\ln \left(2\right)u}{2}\cdot \:2+\ln \left(2\right)u\cdot \:2=\frac{9u^2}{2}\cdot \:2 \\ \\ \ln \left(2\right)u+2\ln \left(2\right)u=9u^2 \\ \\ 3\ln \left(2\right)u=9u^2 \\ \\ 9u^2=3\ln \left(2\right)u \\ \\ 9u^2-3\ln \left(2\right)u=0 \\ \\ u=\frac{-\left(-3\ln \left(2\right)\right)+\sqrt{\left(-3\ln \left(2\right)\right)^2-4\cdot \:9\cdot \:0}}{2\cdot \:9}$$=\frac{3\ln \left(2\right)+\sqrt{\left(-3\ln \left(2\right)\right)^2-4\cdot \:9\cdot \:0}}{2\cdot \:9} \\ \\ =\frac{3\ln \left(2\right)+\sqrt{\left(-3\ln \left(2\right)\right)^2-0\cdot \:4\cdot \:9}}{18} \\ \\ \ln \left(2\right)+\sqrt{\left(-3\ln \left(2\right)\right)^2-4\cdot \:9\cdot \:0} \\ \\ =\frac{6\ln \left(2\right)}{18} \\ \\ =\frac{\ln \left(2\right)}{3}$$=\frac{3\ln \left(2\right)-\sqrt{\left(-3\ln \left(2\right)\right)^2-4\cdot \:9\cdot \:0}}{2\cdot \:9} \\ \\ =\frac{3\ln \left(2\right)-\sqrt{\left(-3\ln \left(2\right)\right)^2-0\cdot \:4\cdot \:9}}{18} \\ \\ 3\ln \left(2\right)-\sqrt{\left(-3\ln \left(2\right)\right)^2-4\cdot \:9\cdot \:0} \\ \\ \sqrt{\left(-3\ln \left(2\right)\right)^2-4\cdot \:9\cdot \:0}=\sqrt{\left(3\ln \left(2\right)\right)^2-0} \\ \\ =\sqrt{\left(3\ln \left(2\right)\right)^2} \\ \\ =3\ln \left(2\right)$$=3\ln \left(2\right)-3\ln \left(2\right)=0=\frac{0}{18}=0 \\ \\ u=\frac{\ln \left(2\right)}{3},\:u=0 \\ \\ u=\frac{\ln \left(2\right)}{3} \\ \\ \ln \left(x\right)=\frac{\ln \left(2\right)}{3} \\ \ln \left(x\right)=\frac{\ln \left(2\right)}{3} \\ \ln \left(x\right)=\ln \left(e^{\frac{\ln \left(2\right)}{3}}\right) \\ x=e^{\frac{\ln \left(2\right)}{3}}$