Exponential Equations

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

To solve such equation we use following property: exponential equation `a^(f(x))=a^(g(x))` is equivalent to the equation `f(x)=g(x)`.

There are two methods of solving of exponential equation:

  1. Method of equating exponents, i.e. transformation of given equation into the form `a^(f(x))=a^(g(x))` and then to the equivalent form `f(x)=g(x)`.
  2. Method of introducing new variable.

Example 1. Solve equation `((0.2)^(x-0.5))/(sqrt(5))=5*(0.04)^(x-1)`.

Let's transform all powers to the base 0.2: we have that `sqrt(5)=sqrt((1/5)^(-1))=sqrt((0.2)^(-1))=(0.2)^(-0.5)`, `5=(0.2)^(-1)`, `(0.04)^(x-1)=((0.2)^2)^(x-1)=(0.2)^(2(x-1))`.

Therefore, equation can be rewritten as `((0.2)^(x-0.5))/((0.2)^(-0.5))=(0.2)^(-1)*(0.2)^(2(x-1))` or `(0.2)^(x-0.5-(-0.5))=(0.2)^(-1+2(x-1))`.

After simplifying we obtain that `(0.2)^x=(0.2)^(2x-3)`. This equation is equivalent to the equation `x=2x-3`, from which `x=3`.

Example 2. Solve equation `4^x+2^(x+1)-24=0`.

Since `4^x=(2^2)^x=(2^x)^2` and `2^(x+1)=2*2^x` then equation can be rewritten as `(2^x)^2+2*2^x-24=0`.

Now we introduce new variable: let `y=2^x` then `y^2+2y-24=0`. This equation has two roots: y=4 and y=-6.

Thus, we obtained set of equations: `2^x=4,\ 2^x=-6`.

From first equation `x=2`, second equation doesn't have roots, because `2^x>0` for any x.

Therefore, initial equation has only one root: x=2.