Rules of logarithms
Quick guide
This is a summary of the rules of logarithms material used in the xlf presentation series. It include numerical examples, and keystrokes for the Hewlett Packard 12C financial calculator.
- \(log(x)\) normally represents the logarithm with base \(e\)
- e's value is \(2.718281828459045\) to 16 digits, or \(2.71828182845904\) at the Excel maximum of 15 digits
- Using the binomial expansion \(e\) is calculated as $$e=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n} \right)^n \ $$ or $$e=\sum_{n=0}^{\infty}\left(\frac{1}{n!} \right) \ $$
RPN keystrokes ►
Logarithms
In the expression \(x^j\), \(j\) is the exponent and \(x\) is the base.
Log rule | Numerical example |
---|---|
\(x^0=1\) | \(5^0=1\) \(\text{12c: }5\;[\text{Enter}]\;0\;[y^x]\;\text{returns} \;1\) |
\(x^1=x\) | \(4^1=4\) \(\text{12c: }4\;[\text{Enter}]\;1\;[y^x]\;\text{returns} \;4\) |
\(x^{-1}=1/x\) | \(2^{-1}=1/2\) \(\text{12c: }2\;[\text{Enter}]\;1\;[\text{CHS}]\;[y^x]\;\text{returns} \;0.5\) |
Product rule \(x^j x^k=x^{j+k}\) | \(x^2x^4=x^{2+4}=x^6\) \(2^2\cdot2^4=4\cdot16=64; \quad 2^6=64\) \(\text{12c: }2\;[\text{Enter}]\;2\;[y^x]\;2\;[\text{Enter}]\;4\;[y^x]\;[×]\;\text{returns}\;64\) |
Quotient rule \(x^j/x^k=x^{j-k}\) | \(x^7/x^3=x^{7-3}=x^4\) \(5^7/5^3=5^{7-3}=5^4=625; \quad 78125/125=625\) \(\text{12c: }5\;[\text{Enter}]\;7\;[y^x]\;5\;[\text{Enter}]\;3\;[y^x]\;[÷]\;\text{returns} \;625\) |
Power rule \((x^j)^k=x^{jk}\) | \((x^2)^4=x^{2\cdot4}=x^8\) \((4^2)^4=4^8=65536; \quad 16^4=65536\) \(\text{12c: }4\;[\text{Enter}]\;2\;[y^x]\;\;4\;[y^x]\;\text{returns} \;65536\) |
\((xy)^j=x^jy^j\) | \((xy)^3=x^3y^3\) \((2\cdot3)^3=2^3\cdot3^3=8\cdot27=216; \quad 6^3=216\) \(\text{12c: }2\;[\text{Enter}]\;3\;[×]\;3\;[y^x]\;\text{returns}\;216\) |
\((x/y)^j=x^j/y^j\) | \((x/y)^2=x^2/y^2\) \((4/5)^2=4^2/5^2=16/25=0.64; \quad(0.8)^2=0.64\) \(\text{12c: }4\;[\text{Enter}]\;5\;[÷]\;2\;[y^x]\;\text{returns}\;0.64\) |
\(x^{-j}=1/x^j\) | \(x^{-3}=1/x^3\) \(2^{-3}=1/2^3=1/8; \quad2^{-3}=0.125\) \(\text{12c: }2\;[\text{Enter}]\;3\;[\text{CHS}]\;[y^x]\;\text{returns}\;0.125\) |
Note 1: \(\dfrac{x^j} {x^j}=1\), thus by application of the quotient rule \(x^{j-j}=x^0=1\)
References
Imperial College London, Euler's Identity' Divide and Rule, Stories from the Complex Plane (page 2), accessed 30 January 2020
- Published: 2 February 2015
- Revised: Thursday 30th of January 2020 - 04:18 PM, [Australian Eastern Standard Time (EST)]