# Logarithmic Equations

Equations of the form log_a(f(x))=log_a(g(x)), where a>0,a!=1 are called logarithmic equations.

To solve such equation we use following property: logarithmic equation log_a(f(x))=log_a(g(x)) has equation-consequence f(x)=g(x).

There are two methods of solving of exponential equation:

1. Method of equating expressions under logarithms, i.e. transformation of given equation into the form log_a(f(x))=log_a(g(x)) and then to the form f(x)=g(x).
2. Method of introducing new variable.

Example 1. Solve equation log_3(x+4)+log_3(2x+3)=log_3(1-2x).

Domain of the equation is {(x+4>0),(2x+3>0),(1-2x>0):} or -3/2<x<1/2.

Next, sum of logarithms equals logarithm of product, therefore log_3((x+4)(2x+3))=log_3(1-2x).

So, we have that (x+4)(2x+3)=(1-2x) or 2x^2+11x+12=1-2x.

After simplifying we obtain that 2x^2+13x+11=0. This equation has two roots: x=-1,x=-11/2, but only x=-1 belongs to the domain of the equation. Therefore, initial equation has one root: x=-1, root x=-11/2 is extraneous.

Example 2. Solve equation log_2^2(x)+log_2(x)+1=7/(log_2(0.5x)).

Domain of the equation is x>0.

Since log_2(0.5x)=log_(2)(0.5)+log_2(x)=-1+log_2(x) then equation can be rewritten as log_2^2(x)+log_2(x)+1=7/(log_2(x)-1).

Now we introduce new variable: let y=log_2(x) then y^2+y+1=7/(y-1). Multiplying both sides by (y-1) gives (y-1)(y^2+y+1)=7 or y^3-1=7. This equation has one root: y=2.

Thus, we obtained equation log_2(x)=2 or x=4. This root belongs to the domain of the equation.

Therefore, initial equation has one root: x=4.