Talk:Algebra II/Logs

Guy's Scratchwork

When writing alone on Wikiversity I like to temporarily store fragments of the article on the talk page.--Guy vandegrift (discuss • contribs) 03:54, 30 November 2021 (UTC)

Appendices

Power to a power rule: $(b^{m})^{n}=b^{mn}\Rightarrow n\log _{b}(x^{n})=n\log _{b}x$

This equation has two exponents, $b^{m}$ and $b^{mn}.$ We need to convert each equation into one that involves a logarithm. In order to do this we need to define two variables, $x$ and $y:$ $x=b^{m}\quad {\text{ and }}\quad y=b^{mn}=x^{n}$. This last equation for $y$ used the Power to a Power rule for exponents. Now we use our prescription for converting equations with powers into equations involving logarithms: $m=\log _{b}(x)\quad {\text{ and }}\quad mn=\log _{b}(y)=\log _{b}(x^{n})$ These two equations can be combined to yield the desired identity: $nm=\underbrace {n\cdot \overbrace {\log _{b}(x)} ^{m}=\log _{b}(x^{n})} _{\text{desired result}}$ Removing all the clutter in this equation yields an important rule concerning the logarithm of $x$ raised to the $n^{\text{th}}$ power: $n\log _{b}x=log_{b}x^{n}$|} The other identities need to proven in the same fashion. Among them are: $b^{m}/b^{n}=b^{m-n}\Rightarrow \log _{b}\left({\frac {m}{n}}\right)=log_{b}(m)-log_{b}(n)$. Leave a message at User talk:Guy vandegrift if you are a student, professor or other scholar interested proving this or related identities. This is a Quizbank project. Quotient of powers (exponents): $b^{m}/b^{n}=b^{m-n}\Rightarrow \log _{b}\left({\frac {m}{n}}\right)=log_{b}(m)-log_{b}(n)$ Scratchwork {{NumBlk|:|<math>xx   </math>|n|LnSty=.1px dashed red}} <!--Do not change title (incoming links)-->