An introduction to exponential equation

We have the sufficient skills to be able to solve problems of the type $$a^b=c$$, where $$a,b$$ and $$c$$ are real numbers, and $$c$$ has to be determined.

We want to know how much $$3^3$$ is, which means to multiply $$3$$ by itself $$3$$ times, so it is $$3^3=3 \cdot 3\cdot 3=27$$

These terms are called potencies.

Now the same type of calculations is going to applied to solve potency, but with the term $$b$$ to be determined, that is to say, when the unknown is in the exponent (which is called an exponential function).

An exponential equation is characterized by having in some of its members an exponential function. Therefore, to solve them one needs a solid basis in the solution of expressions of the exponential type, that is to say, for example:

$$$5^a=8 \Rightarrow \log_5 5^a=\log_5 8 \Rightarrow a= \log_5 8$$$ Let's remember it step by step:

We want to find $$a$$ so that: $$$6^a=24$$$ We apply logarithms, first to the left of the equality, so we have: $$$\log_6 6^a= a \cdot \log_6 6=a \cdot 1=a$$$ While on the right we have: $$$\log_6 24= \log_6(6\cdot 4)=\log_56+\log_6 4=1+\log_64$$$ Therefore, our equality now is: $$$a= 1+\log_6 4$$$

As has been said, the exponential function is the one in which the variable is the exponent of a power. Namely, for example, an equality of the type: $$a^x=b$$ where $$a$$ and they $$b$$ are known and $$x$$ is the variable to be determined. These use basic knowledge of the exponential and logarithmic functions.

In an expression of the type $$a^x=b$$ the following names are assigned to the appearing elements:

  • $$a$$ is the base of the potency
  • $$x$$ is the exponent
  • $$b$$ is the potency

Its solution, as it has been previously mentioned is $$x=\log_a b$$ where the elements that appear are designated by:

  • $$x$$ is the logarithm
  • $$a$$ is the base of the logarithm
  • $$b$$ is the number of which we calculate the logarithm


  • $$a^0=1$$ for any $$a$$
  • For any $$a \neq 0$$ and $$a \neq 1$$ we have: $$$a^x=b \Longrightarrow x= \log_a b$$$

Proceeding by means of these two rules, and with knowledge of solving the operations, combined equations can already be solved with terms including an exponential function. See an example:

$$$\displaystyle \frac{(17+3)}{2}+3 \cdot 5^x-7=\frac{12}{2^2}+3$$$

The hierarchy of the combined operations is ruled by the following order.

  • First: calculate whatever is in brackets, square brackets and keys.
  • Second: the power and roots are calculated.
  • Third: the products and quotients are performed.
  • Quarter: the sums and the subtractions are done.

Therefore, the brackets are solved first $$$(17+3)=20$$$ Re-writing we obtain $$$\displaystyle \frac{(17+3)}{2}+3 \cdot 5^x-7=\frac{12}{2^2}+3$$$ The poweris calculated, except that of $$5^x$$ since there is the unknown to be determined. We have $$2^2=4$$ left, so we have: $$$\displaystyle \frac{20}{2}+3 \cdot 5^x-7=\frac{12}{4}+3$$$ The products and quotients are calculated $$$10+3^3 \cdot 5^x-7=3+3$$$ Performing the sums and subtractions we obtain: $$$3 \cdot 5^x=3+3+7-10=3 \Longrightarrow 5^x=\displaystyle \frac{3}{3}=1$$$ Therefore the result is $$$x=\log_5 1$$$

To create an exercise of this type it is necessary to bear in mind that if we want the solution to exist, all the logarithms that appear must be positive numbers. This is because there exists no $$x$$ number such that $$a^x=0$$ since, if we apply logarithms, $$x=\log 0$$ cannot exist either.

That's why, when one wants an equation with a solution, a procedure that assures this involves starting with the solution and doing it in the inverse way. That is, for example:

$$$\displaystyle x=\log_3 21 \\ 3^x=3^{\log_3 21}=21$$$

We already have an exponential equation whose solution is: $$3^x=21$$ .

If one wants to make it more complicated, it can be complemeted with, for example: $$$3^x=21 \Rightarrow 3^x=50-21=2\cdot 5^2-(7 \cdot +2^3) \Rightarrow 3^x-2 \cdot 5^2=-(7 \cdot 3+2^3)$$$ which keeps on being the same, although it needs a few first additional steps to calculate combined operations since we have only decomposed $$21$$ into the difference between two numbers, which are latterly broken down into product of prime numbers.