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.