Natural Logarithms and Exponentials

We assume values of natural logarithms and and its inverse, the exponential function, can be obtained from electronic calculators. How will be described. The properties of natural logarithms and its inverse, the exponential function, provide a framework for the calculation of of further logarithms and exponentials in this lesson, and for the calculation of roots and powers with fractional and real exponents in following lessons. All the foregoing provide a base for and several interchangeable, that is equivalent, growth and decay models. The explanations here are precalculus

Domain - Where Defined.

• The natural logarithm $\ln(x)$ is defined for x > 0.

• The exponential function $\exp(x)$ is defined for all real x.
  

With a calculator try to find the values of

\begin{eqnarray*} A&=& \exp(0) \\ B&=& \ln(2.718281828) \\ C&=& \ln(-3) \end{eqnarray*}

Results should be 0, 1 or close to, and error [or undefined]

With a calculator use the $e^x$ button to calculate

\begin{eqnarray*} D&=& \ln(8) \\ E&=& \exp(1) \\ F&=& \exp(x) \mbox{ when } x=\ln(5) \end{eqnarray*}

Results should be 1, 2.718281828 approx, and 5. The relation between $e^x$ and $exp(x)$ will be explained later.

Uniqueness [1 to 1] Property:

If a > 0, b > 0 and $\ln(a) = \ln(b)$ then a = b.

In other words, if $\ln(a)$ and $ln(b)$ have the same value for two positve numbers [or expressions with numerical values] a and b then the two numbers must be equal.

Inversion Properties

\begin{eqnarray*} \ln(\exp(x)) &= x & \quad \mbox{for all real x} \\ \exp(\ln(x)) &= x& \quad \mbox{ if } x \gt 0 \end{eqnarray*}

Aside: For each real number a, \[x = \exp(a) \quad\mbox{the unique solution of }\quad a = \ln(x)\]
Solving the latter for x graphically is one way to define or compute $\exp(a).$

Fundamental property of logarithms

Caution: the capital EXP on some calculators will not help you with the calculation of exp(x). Use the button marked e x instead.

For all positive numbers a and b,

\[\ln(ab) = \ln(a) + \ln(b) ) \]

The reasons why or proof available in calculus - see chapter 19 of Volume 3, Why Slopes and More Math.

Fundamental property of exponentials

For all real numbers c and d,

\[\exp(c+d) = \exp(c)\cdot \exp(d) \]

This follows algebraically from the fundamental properties of logarithms combined with the uniqueness property of logarithms with the inversion properties of the exponential function.

Reciprocal or Multiplicative Inverse Properties

The fundamental property of logarithms implies

\[ \ln( \frac1a) = (-1) \ln(a) \]

That is because \[ 0 = \ln(1) = \ln ( \frac1a \cdot a ) = \ln(\frac1a)+ \ln(a) \]

In the case of a nonzero fraction $a = \frac{q}{p},$ the multiplicative inverse and reciprocal are both equal to \[\frac 1a = \frac {p}{q} = 1 \div a \] The multiplicative inverse and reciprocal of $4$ is $\frac14 = 1 \div 4 = 0.25$ while the multiplicative inverse and reciprocal of $\pi$ is $\frac1\pi = 1 \div \pi.$

The property $\exp(0) = 1$ with

\[\exp(c+d) = \exp(c)\cdot \exp(d) \]

\begin{eqnarray*} \exp(-a+a) &=& \exp(-a)\cdot \exp(a) \\ \hbox {or} \\ 1 &=& \exp(-a)\cdot \exp(a) \end{eqnarray*} Whence \[\exp(-a) = \exp(a) \] for all real numbers a.

A Simple Exercise:

With a calculator find \begin{eqnarray*} A &=&ln(1) \\ B &=&\ln(2) \\ C &=&ln(\frac12) \\ D &=&ln(4) \\ E &=& \ln(0.25) \end{eqnarray*}

Results should have the form \begin{eqnarray*}A &=&\ln(1) =0 \\ B &=&\ln(2) \approx 0.69315 \\ C &=&\ln(\frac12) = -\ln(2) \\ D &=&\ln(4)= 2\ln(2) \\ E &=& \ln(0.25)=-2\ln(2) \end{eqnarray*}

The relation between these values need to be explained.

Powers with whole number exponents

\begin{eqnarray*} f_a(1) &=& a^1 \\ f_a(2) &= &a^2 \\ f_a(3) &=& a^3 \\ f_a (4) &=& a^4 \end{eqnarray*}

or demonstrating these patterns numerically suggests the general pattern

\[ \exp(m \ln(a))=f_a(m) = a^m\]

\[a^m = \exp(m \ln(a))\]

for whole numbers m. The property $\exp(-x) = 1/exp(x)$ then implies

Natural Logarithm of Powers

\[\ln(a^m) = m \cdot\ln (a)\]

for all whole numbers m and then for all integers m. The proof follow from mathematical induction and the reciprocal property of logarithms.

Here is an easy consequence. Apply the anti-log or inverse natural logarithm function to both sides to get

\[\exp( \ln(a^m) ) = \exp( m \cdot\ln (a))\]

or

\[ a^m = \exp( m \cdot\ln (a))\]

due to the inverse logarithm property:

\[ \exp(\ln(x)) = x\]

Example: Calculate and compare the values of \[ (1.7)^3 \mbox{ and }\exp(3 \ln(1.7))\] with the aid of a calculator. Do likewise with the values of \[ 2^5 = 2 \times 2 \times 2 \times 2 \times 2 \] [five equal factors] and the expression\[\exp(5 \ln(2))\]

Calculations should provide numerical confirmation of the equality - if not redo them.

Logarithms to base c > 0.

The logarithm of x > 0 to a base c > 0 is given by

Thus the natural logarithm and logarithm to base e are identical.

The common logarithm of x > 0 to a base 10 is given by

In years gone by, the common logarithm to base 10 was used in place of the more technical logarithm, the natural logarithm $\ln(x)$. Since the advent of calculators and the advent of more people masterying calculus, the natural logarithm has gained in favour. Whether you know it or not, if you are in course that covers logarithms, you are in a course whose design is motivated by calculus. The early introduction of the natural logarithm is motivated by calculus and preparation for it.

Inherited Properties of $\log_c(x)$

For $c \gt 0,$ the definition of $\log_c(x)$ in terms of $\ln(x)$ very easily gives or implies

\begin{eqnarray*} \log_c(ab)&=& \log_c(a)+\log_c(b) \\ \\ \log_c\left(\frac1a\right)&=& - \log_c(a) \\ \\ \log_c(a^m) &=& m \cdot\log_c (a) \end{eqnarray*}

when a and b are both positive and m is an integer.

Exercise: Use the definition of and the fundamental properties of the natural logarithm to show why.

Backward Use of $A = P(1+i)^m$ - Derivation of formula for $m$

In the compound interest, [a.k.a. compound growth and decay] formula $A = P(1+i)^m,$ there may be a question of what is $m$ when quantities A, P and i are known. Here we will employ the properties of the natural logarithm to find n. Steps follow.

• Observe \[ [1+i]^m =\frac AP \]

• Now the natural logarithm of both sides has to be equal \[ \ln\left([1+i]^m\right)=\ln\left(\frac AP\right)\]

• Replace the left hand side by its equal - bring the exponent m outside \[ m\cdot \ln ([1+i] )=\ln\left(\frac AP\right)\]

• Isolate the m by dividing both sides by its coefficient \[ m =\ln\left(\frac AP\right)\frac1{\ln (1+i)} = \frac{\ln(A)-\ln(B)}{\ln (1+i)} \]

The latter expression for m can written in several ways - whatever is the most pleasing according to yourself, or the one correcting your work and giving marks

Definition of Natural Logarithms -
What the $\ln(x)$ button computes for x > 0.

The next two diagram show the area-based definition of the natural logarithm ln(a) or ln(b) in the two mutually exclusive cases a > 1 and 0 < b < 1. Note: The graph of y=ln(x) is different from the graph of t = 1/s. The latter is used to define $\ln(x)$, not graph it.

For a ³ 1, the value of ln(a) is given by the area from s = 1 to s = a under the curve y = [1/(s)]. Here we take or assume ln(1) = 0. It can be shown that ln(a) ® 0 when when a approaches 1 through values above or greater than 1. Observe increasing a increases the area under the curve = ln(a).

For 0 < b < 1, the value of ln(b) is given by (-1) times the area under the curve y = [1/(s)] from s = b to s = 1.

Arithmetic - Ages 10+
1. Deciml Place Value - fun
2. Decimals for Tutors
3. Prime Factors - quickly
4. Fractions + Ratios
5. Arith with units - science