Exponential equations

An ordinary equation becomes an exponential equation when it contains the variable in an index. In general we could write an exponential equation as a^f(x) = b^g(x), where a, b>0. A typical example of an exponential equation is 2^x = 8. Here it is fairly obvious that the answer will be three, because two cubed is eight. Now comes a slightly more mathematical method.

A simple exponential equation

If we want to solve an exponential equation, it is very convenient if we can rearrange it into the form a^f(x) = b^g(x) with a = b. When the bases are the same, the powers are equal exactly when the indices are equal as well. So we only have to solve the equation f(x) = g(x). We can rework the previous example like this: we write eight as a power of two, 2^x = 2³, which gives us the same bases, so we simply equate the two indices and get the answer x = 3. We can make the example harder, for instance like this: 2^2x = 8, and after rewriting we again get 2^2x = 2³ and solve 2x = 3, so the answer is 1.5.

Another simple case occurs when there is a one on one side of the equation. Take this example:

$$5^{3x-2}=1$$

It looks complicated, but if you know the basic rules for powers, you know that there are only two ways to get one: either the base is one, and then one to any power is again one, or the index is zero, because anything to the power zero is again one. We will naturally use the second property and rewrite the equation like this: 5^{3x − 2} = 5⁰. Now we have the equation in a form we already know how to solve (see the previous paragraph). All we have to do is compare the indices. This gives the equation 3x − 2 = 0, which is a trivial linear equation.

Taking logarithms of an exponential equation

If we cannot rewrite the equation with the same bases, we can try taking logarithms. Our maths teacher always used to say “Kids, if you don't know what to do next, just take logs!”, and she was right. If you have an exponential equation with different bases, and it is not possible (or not efficient) to rewrite them with the same base, take the logarithm of the whole equation. From the original equation

$$a^{f(x)}=b^{g(x)}$$

we get

$$\log a^{f(x)}=\log b^{g(x)}$$

and, using the rule for working with logarithms, we get

$$f(x)\cdot \log a=g(x)\cdot \log b.$$

Here log means log₁₀. Let's take this example

$$2^x\cdot 5^{2x}=3^{x-2}$$

The bases are not the same and cannot be made the same, so we take the logarithm of the whole equation:

$$\log (2^x\cdot 5^{2x})=\log (3^{x-2})$$

Now we use the laws of logarithms and “expand” the first bracket

$$\log 2^x+\log 5^{2x}=\log 3^{x-2}$$

Again we use the laws of logarithms and “move” the index in front of the logarithm.

$$x\cdot \log 2+2x\cdot \log 5=(x-2)\log 3$$

We expand the brackets on the right-hand side of the equation:

$$x\cdot \log 2+2x\cdot \log 5=x\log 3-2\log 3$$

We move the terms with the unknown to the left-hand side of the equation:

$$x\cdot \log 2+2x\cdot \log 5-x\cdot \log 3=-2\log 3$$

Next we take out the common factor x:

$$x(\log 2+2\log 5-\log 3)=-2\log 3$$

We make x the subject:

$$x=\frac{-2\log 3}{\log 2+2\log 5-\log 3}$$

That is more or less the answer. We could apply the laws of logarithms once more and turn 2log 5 into log 5² = log 25, but that really is the last thing we can do with the result.

Substitution

We can also solve exponential equations using substitution. We will show it on an example:

$$7^{2x}+7^x-6=0.$$

Let's substitute a = 7^x. Now we rewrite the original equation, replacing 7^x with a. This gives the following (no longer exponential) equation:

$$a^2+a-6=0.$$

This is a plain quadratic equation. Its roots are the numbers 2 and −3.

Now we substitute these roots into a = 7^x. So 2 = 7^x. We take logarithms:

$$\log 2=x\cdot \log 7$$

and make x the subject:

$$x=\frac{\log 2}{\log 7}$$

We do not need to substitute the second root, because it is negative – we would end up with the logarithm of a negative number, which is not possible. And that is the solution of the exponential equation.

Examples

Solve the following exponential equation:

$$2^{3x-4}=8^{2x+1}.$$

At first glance the bases on the two sides are not equal. Sad. But at second glance we can surely see a step that will give us the same bases. Instead of eight we will simply use 2³, which is equal to eight. Using the rule for a power of a power we get:

$$\begin{array}{rcl}{2}^{3x-4}& =& 8^{2x+1}\\ 2^{3x-4}& =& 2^{3\cdot (2x+1)}\\ 2^{3x-4}& =& 2^{6x+3}\end{array}$$

Now the bases are equal and we can solve the simple linear equation 3x − 4 = 6x + 3:

$$\begin{array}{rcl}3x-4& =& 6x+3\\ -3x& =& 7\\ x& =& -\frac{7}{3}\end{array}$$

Solve the following exponential equation:

$$5^x\cdot 2^x={100}^{x-1}.$$

Here, as usual, we can see that the bases are not the same. But we probably all suspect that they can be converted somehow. On the left-hand side we apply the rule for multiplying powers with the same index, and on the right-hand side we apply the same rule as a moment ago and turn 100^{x − 1} into 10^{2(x − 1)}:

$$\begin{array}{rcl}{5}^x\cdot 2^x& =& {100}^{x-1}\\ {10}^x& =& {10}^{2(x-1)}\\ {10}^x& =& {10}^{2x-2}\end{array}$$

And once again we have the same bases and can solve a classic linear equation:

$$\begin{array}{rcl}x& =& 2x-2\\ -x& =& -2\\ x& =& 2\end{array}$$

Solve the following exponential equation:

$$3^x+3^{x+1}=108.$$

As always, we first have to work our way to the same base. Here taking out a common factor helps: we take 3^x out of the expression on the left-hand side. Then we add up the expression in the bracket and get a simple equation.

$$\begin{array}{rcl}{3}^x+3^{x+1}& =& 108\\ 3^x(1+3^1)& =& 108\\ 4\cdot 3^x& =& 108\\ 3^x& =& 27\end{array}$$

Now we have rewritten the left-hand side, so it is time to rewrite the right-hand side. It is quite clear that this is the third power of three:

$$\begin{array}{rcl}{3}^x& =& 27\\ 3^x& =& 3^3\end{array}$$

And now only the last step remains – the bases are equal, so we equate the indices:

$$x=3$$

Solve the following exponential equation:

$$4^{2x}-6\cdot 4^x+8=0.$$

In this case we try to turn the exponential equation into an ordinary quadratic equation. The best way to get there is the substitution 4^x = a:

$$a^2-6a+8=0$$

Now we have an ordinary quadratic equation, so we work out the discriminant and the roots:

$$\begin{array}{rcl}D& =& 36-4\cdot 8\\ a_{1,2}& =& \frac{(6\pm 2)}{2}\\ a_1& =& 4\\ a_2& =& 2\end{array}$$

We still have to substitute these partial results back into the substitution. The first result:

$$\begin{array}{rcl}{4}^x& =& 4\\ x& =& 1\end{array}$$

The second result:

$$\begin{array}{rcl}{4}^x& =& 2\\ 4^x& =& 4^{\frac{1}{2}}\\ x& =& \frac{1}{2}\end{array}$$