Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 180498 by Spillover last updated on 12/Nov/22

Commented by Frix last updated on 13/Nov/22

$$\mathrm{more}\mathrm{challenging}\mathrm{extra}\mathrm{work}:$$$$\omega =\frac{{(\frac{{\log }_{a^3+b^4}^{2}z}{25}-R^2)}^{\frac{1}{3}}\pm {({\left((\frac{{\log }_{a^3+b^4}^{2}z}{25}-R^2)C\right)}^{\frac{2}{3}}+4L)}^{\frac{1}{2}}}{2C^{\frac{1}{3}}L}$$

Commented by Spillover last updated on 13/Nov/22

$$25\mathrm{where}\mathrm{did}\mathrm{you}\mathrm{get}\mathrm{it}\mathrm{from}@\mathrm{Frix}$$

Commented by Frix last updated on 13/Nov/22

$$\ln z^{\frac{1}{5}}=\sqrt{term}\ln (a^3+b^4)$$$$\frac{\ln z}{5}=\sqrt{term}\ln (a^3+b^4)$$$$\frac{\ln z}{5\ln (a^3+b^4)}=\sqrt{term}$$$$\mathrm{rule}\frac{\ln a}{\ln b}={\log }_{b}a$$$$\frac{{\log }_{a^3+b^4}z}{5}=\sqrt{term}$$$$\frac{{\log }_{a^3+b^4}^{2}z}{25}=term$$$$$$$$\mathrm{I}\mathrm{wrote}\ln \mathrm{instead}\mathrm{of}\log ,\mathrm{corrected}\mathrm{this}\mathrm{now}$$

Answered by Frix last updated on 13/Nov/22

$$C=\frac{1}{\sqrt{{\omega }^3{(\omega L-\sqrt[3]{\frac{{\log }_{a^3+b^4}^{2}z}{25}-R^2})}^3}}$$

Answered by floor(10²Eta[1]) last updated on 13/Nov/22

$${\mathrm{lnz}}^{1/5}=\sqrt{{\mathrm{R}}^2+{(\omega \mathrm{L}-\frac{1}{\omega {\mathrm{C}}^{2/3}})}^3}\ln ({\mathrm{a}}^3+{\mathrm{b}}^4)$$$${\left(\frac{{\mathrm{lnz}}^{1/5}}{\ln ({\mathrm{a}}^3+{\mathrm{b}}^4)}\right)}^2={\mathrm{R}}^2+{(\omega \mathrm{L}-\frac{1}{\omega {\mathrm{C}}^{2/3}})}^3$$$$\sqrt[3]{{\left(\frac{{\mathrm{lnz}}^{1/5}}{\ln ({\mathrm{a}}^3+{\mathrm{b}}^4)}\right)}^2-{\mathrm{R}}^2}=\omega \mathrm{L}-\frac{1}{\omega {\mathrm{C}}^{2/3}}$$$$\frac{1}{\omega {\mathrm{C}}^{2/3}}=\omega \mathrm{L}-\sqrt[3]{{\left(\frac{{\mathrm{lnz}}^{1/5}}{\ln ({\mathrm{a}}^3+{\mathrm{b}}^4)}\right)}^2-{\mathrm{R}}^2}$$$$\frac{1}{{\mathrm{C}}^{2/3}}={\omega }^2\mathrm{L}-\omega \sqrt[3]{{\left(\frac{{\mathrm{lnz}}^{1/5}}{\ln ({\mathrm{a}}^3+{\mathrm{b}}^4)}\right)}^2-{\mathrm{R}}^2}$$$${\mathrm{C}}^{2/3}=\frac{1}{{\omega }^2\mathrm{L}-\omega \sqrt[3]{{\left(\frac{{\mathrm{lnz}}^{1/5}}{\ln ({\mathrm{a}}^3+{\mathrm{b}}^4)}\right)}^2-{\mathrm{R}}^2}}$$$$\mathrm{C}=\frac{1}{\sqrt{{({\omega }^2\mathrm{L}-\omega \sqrt[3]{{\left(\frac{{\mathrm{lnz}}^{1/5}}{\ln ({\mathrm{a}}^3+{\mathrm{b}}^4)}\right)}^2-{\mathrm{R}}^2})}^3}}$$$$$$

Commented by Spillover last updated on 13/Nov/22

$$\mathrm{great}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com