Question and Answers Forum

All Questions      Topic List

Logarithms Questions

Previous in All Question      Next in All Question      

Previous in Logarithms      Next in Logarithms      

Question Number 91681 by hmamarques1994@gmail.com last updated on 02/May/20

$$$$$$\mathbf{Se}\mathbf{f}(\sqrt{3^{\sqrt[3]{\mathbf{x}}}}+3^{\sqrt[3]{\mathbf{x}}})=\sqrt[3]{\mathbf{x}},\mathbf{calcule}\frac{\mathbf{f}\left(2\right)}{\mathbf{f}\left(1\right)}\cdot$$$$$$

Commented by john santu last updated on 02/May/20

$$set\sqrt[3]{x}=t$$$$f(3^{\frac{t}{2}}+3^t)=t$$$$\left(1\right)f\left(2\right);2=3^{\frac{t}{2}}+3^t,{\left(3^{\frac{t}{2}}\right)}^2+3^{\frac{t}{2}}-2=0$$$$3^{\frac{t}{2}}=1\Rightarrow t=0,f\left(2\right)=0$$$$\left(2\right)f\left(1\right);{\left(3^{\frac{t}{2}}\right)}^2+3^{\frac{t}{2}}-1=0$$$$3^{\frac{t}{2}}=\frac{-1+\sqrt{5}}{2}\Rightarrow \frac{t}{2}={\log }_3\left(\frac{\sqrt{5}-1}{2}\right)$$$$t={\log }_3\left(\frac{3-\sqrt{5}}{2}\right)\Rightarrow f\left(1\right)={\log }_3\left(\frac{3-\sqrt{5}}{2}\right)$$$$\left(3\right)\frac{f\left(2\right)}{f\left(1\right)}=0$$

Commented by hmamarques1994@gmail.com last updated on 02/May/20

$$Show,man!$$

Commented by john santu last updated on 02/May/20

$$manshow$$

Answered by mr W last updated on 02/May/20

$$sayt=\sqrt{3^{\sqrt[3]{x}}}>0$$$$t^2=3^{\sqrt[3]{x}}$$$$f(t+t^2)={\log }_3{t}^2$$$$sayu=t+t^2$$$$t^2+t-u=0$$$$t=\frac{\sqrt{1+4u}-1}{2}$$$$t^2=u-t=u-\frac{\sqrt{1+4u}-1}{2}=\frac{2u+1-\sqrt{1+4u}}{2}$$$$f\left(u\right)={\log }_3\frac{2u+1-\sqrt{1+4u}}{2}$$$$\Rightarrow f\left(x\right)={\log }_3\frac{2x+1-\sqrt{1+4x}}{2}$$$$\Rightarrow \frac{f\left(2\right)}{f\left(1\right)}=\frac{{\log }_3\frac{5-3}{2}}{{\log }_3\frac{3-\sqrt{5}}{2}}=\frac{0}{{\log }_3\frac{3-\sqrt{5}}{2}}=0$$

Commented by hmamarques1994@gmail.com last updated on 02/May/20

$$\mathit{S}\mathit{h}\mathit{o}\mathit{w},\mathit{m}\mathit{a}\mathit{n}!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com