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
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): $$b^m/b^n=b^{m-n}\Rightarrow \log_b\left(\frac m n\right) = log_b( m) - log_b( n)$$