# Logarithmic Functions Since exponential function y=a^x is monotonic (i.e. either increasing or decreasing) then it passes horizontal line test. Thus, it has inverse function. To find this function we use algorithm for finding inverse.

Interchange x and y: x=a^y. To write this in terms of y we introduce new record: y=log_a(x).

So, x=a^y is equivalent to y=log_a(x). For example, log_2(8)=3 because 2^3=8.

Function y=log_a(x) is called logarithmic function. Its domain is (0,oo) and range is (-oo,oo).

Since y=log_a(x) is inverse of y=a^x then their graphs are symmetric about line y=x.

Properties of Logarithms

If x and y are positive numbers then

1. a^(log_a(x))=x, for any x,
2. log_a(a^x)=x, for x>0,
3. log_a(xy)=log_a(x)+log_a(y),
4. log_a(x/y)=log_a(x)-log_a(y),
5. log_a(x^b)=blog_a(x) (where b is any real number)

First two properties follow from definition of inverse, last three properties follow from corresponding properties of exponentials.

Example 1. Evaluate log_2(3)+log_2(20)-log_2(15).

Using properties of logarithms we can write that log_2(3)+log_2(20)-log_2(15)=log_2(3*20)-log_2(15)=log_2(60)-log_2(15)=log_2(60/15)=log_2(4)=2

because 2^2=4.

Natural logarithm

Of all possible bases a the most convenient when working with logarithms is base e (number e~~2.718). We will talk why is this so in next notes.

So, logarithm with base e is called natural logarithm and is denoted as ln(x): log_e(x)=ln(x).

From definition of inverse it follows that ln(x)=y\ <=>\ e^y=x.

Also, ln(e^x)=x for x in (-oo,\ oo) and e^(ln(x))=x for x>0.

In particular ln(e)=1.

Example 2. Solve equation e^(3x+5)=4.

Take natural logaritms of both sides: ln(e^(3x+5))=ln(4) or 3x+5=ln(4) , so x=(ln(4)-5)/3.

Example 3. Solve ln(x^3+1)=2.

Apply exponential function to both sides of equation: e^(ln(x^3+1))=e^2 or x^3+1=e^2.

So, x=root(3)(e^2-1).

Example 4. Simplify 2ln(x)-1/3 ln(y).

Using properties of logarithms we can write that 2ln(x)-1/3 ln(y)=ln(x^2)-ln(y^(1/3))=ln((x^2)/(y^(1/3))).

Now, let's see how can we express logaritm with one base through logarithm with another base.

Let y=log_a(x). This means that a^y=x.

Taking logarithms with base b we have that log_b(a^y)=log_b(x) or y log_b(a)=log_b(x).

Thus, y=(log_b(x))/(log_b(a)). From another side y=log_a(x).

Finally, log_a(x)=1/(log_b(a)) log_b(x).

Change of base formula. log_a(x)=(log_b(x))/(log_b(a)).

In case when b=e we can write equation as color(blue)(log_a(x)=ln(x)/ln(a)).