More About this Exponential Equation Calculator

The main purpose of this calculator is to solve exponential equations that you provide, and to show you the solution involving all the steps. For example, you may type an equation such as '9^x + 3^x = 4'.

Once you are satisfied with the equation you have typed, then you go and click on "Solve", so the the steps of the solution is provided, with all the steps involved.

Exponential equations are usually solved by making use of some of the different Laws of Exponents.

What is an Exponential Equation

An exponential equation is, in simple terms, an algebra equation in which the unknown (x), appears as an exponent. For example,

\[\displaystyle 2^x = 4 \]

is a simple exponential equation, because the unknown variable we want to solve for (x), appears as an exponent, with base 2. Now, you have more complicated exponential equations, such as the example below:

\[\displaystyle \cos(2^x) + e^x = 4x \]

What are the steps for solving exponential equations

• Step 1: Make sure that you are dealing with an exponential equation, for which you need to see whether x appears as an exponent
• Step 2: It is important to ensure you are working with an exponential equation. If not, you will likely have to use a different approach
• Step 3: Be aware that not all exponential equations you encounter will be easy to solve, or even you may not be able to solve it
• Step 4: The main strategy would be to try to group all exponential parts in one exponential expression, if possible. For example, if you have an equation like \(2^x 2^y = 4\), you will want rewrite it as \(2^{x+y} = 2^2\)
• Step 5: Put all that depends on x (and all the unknowns) on one side and the rest on the other side
• Step 6: Then, you are trying to put together all exponential parts in one, so that you will try to equate exponents

The main idea is to group exponents as much as possible so that, as you can imagine, we can eliminate the base. So in other words, the strategy for solving exponential equation is to actually getting rid of the exponential part of it.

How do you find the exponential equation?

Exponential equations appear naturally in different Algebra settings. For example they are very common when dealing with population models and growth rates, or when dealing with application problems about radioactive decay and half life.

Typically, the context will dictate what type of base and exponent you will find or you will use when solving an exponential equation. For example, you may have the case of a situation of some microorganism that starting duplicating itself every hour, and you would like to know how many hours until the microorganism population reaches 1,000,000.

In this context, it is not hard to realize that the population after \(x\) hours is \(2^x\), and then from the setting of the problem, we want to solve the equation:

\[\displaystyle 2^x = 1,000,000 \]

What are basic uses of exponential equations?

• Use 1: Modeling population growth based on exponential growth
• Use 2: Modeling exponential decay and computing half-life, for example that exhibited by radioactive materials
• Use 3: Financial applications for continuous compounding

The main ideas in Algebra linked to exponential equations are exponential growth in decay which are observed in the examples detailed above.

How do you find an exponential function with two points?

Exponential functions are important because they are the main components found in an exponential equation. You can use this exponential function calculator to find the function from two points.

There are other ways to determine exponential function, namely, using the approach of the initial value and growth rate, in which case you can use the same calculator from the link above.

It is certainly useful to have an expontial equation calculator with steps, so to eliminate the guesswork of what needs to be done in order to solve the equation, though often times you will find out that not all equations can be solved with the methods we know.

Example: Calculating a simple exponential equation

Solve: \(2^{2x+1} = 4\)

Solution: The following equation needs to be solved:

\[2^{2x+1}=4\]

We observe that:

Putting \(x\) on the left hand side and the constant on the right hand side we get

\[\displaystyle 2x = 1\]

Then, solving for \(x\), by dividing both sides of the equation by \(2\), the following is obtained

\[\displaystyle x=\frac{1}{2}\]

Therefore, we find that that the auxiliary equation has one real solution, which is: \(x = \frac{1}{2}\)

Plugging this value back into the original equation confirms that this is a solution. which concludes the calculation.

Example: Solving exponential equations via substitution

Solve the the following: \(9^x + 3^x = 4\)

Solution: We have the following equation:

\[9^x+3^x=4\]

So then:

By solving this rational equation in the variable \(u\), and then using that \(u = 3^x\), we get the solutions \[x_1=\frac{2i\pi{}K_1+\ln\left(\frac{1}{2}\sqrt{17}-\frac{1}{2}\right)}{\ln\left(3\right)} \text{ , for } K_1 \text{ an arbitrary integer constant}\] \[ = ... \, \frac{-2i\pi{}+\ln\left(\frac{1}{2}\sqrt{17}-\frac{1}{2}\right)}{\ln\left(3\right)}, \, \,\, \frac{\ln\left(\frac{1}{2}\sqrt{17}-\frac{1}{2}\right)}{\ln\left(3\right)}, \,\, \, \frac{2i\pi{}+\ln\left(\frac{1}{2}\sqrt{17}-\frac{1}{2}\right)}{\ln\left(3\right)}, \, \, \, \frac{4i\pi{}+\ln\left(\frac{1}{2}\sqrt{17}-\frac{1}{2}\right)}{\ln\left(3\right)} \, ...\]
\[x_2=\frac{i\pi{}+2i\pi{}K_2+\ln\left(\frac{1}{2}\sqrt{17}+\frac{1}{2}\right)}{\ln\left(3\right)} \text{ , for } K_2 \text{ an arbitrary integer constant}\] \[ = ... \, \frac{-i\pi{}+\ln\left(\frac{1}{2}\sqrt{17}+\frac{1}{2}\right)}{\ln\left(3\right)}, \, \,\, \frac{i\pi{}+\ln\left(\frac{1}{2}\sqrt{17}+\frac{1}{2}\right)}{\ln\left(3\right)}, \,\, \, \frac{3i\pi{}+\ln\left(\frac{1}{2}\sqrt{17}+\frac{1}{2}\right)}{\ln\left(3\right)}, \, \, \, \frac{5i\pi{}+\ln\left(\frac{1}{2}\sqrt{17}+\frac{1}{2}\right)}{\ln\left(3\right)} \, ...\]

Therefore, solving for \(x\) for the given equation leads to the solutions \(x=\frac{2i\pi{}K_1+\ln\left(\frac{1}{2}\sqrt{17}-\frac{1}{2}\right)}{\ln\left(3\right)},\,\,x=\frac{i\pi{}+2i\pi{}K_2+\ln\left(\frac{1}{2}\sqrt{17}+\frac{1}{2}\right)}{\ln\left(3\right)}\), for \(K_1, K_2\) arbitrary integer constants.

Real Solutions

The given equation has been found to have both complex and real solutions. The real solution identified is \(x=\frac{\ln\left(\frac{1}{2}\sqrt{17}-\frac{1}{2}\right)}{\ln\left(3\right)}\).

More equation calculators

Other related operations you may want to do is to solve quadratic equations, or solve a linear equation, which in the grand scheme of things, are the easiest to solve and guaranteed to find all the solutions.

Then you can also use a trigonometric equation solver, to deal with those often tricky trig equations that pop up from time to time.

By using an equation calculator like the ones referred to you will see clearly wow do you solve the equation, and if an equation cannot be solved, where is the point we realize that, or just why we cannot do it.