Exponential equations

An ordinary equation becomes exponential if it contains a variable in the exponent. In general, we might write an exponential equation like this: $a^{f(x)} = b^{g(x)}$, where a, b>0. A typical example of an exponential equation might be 2^x = 8. Here it is quite obvious that the result will be the number three, because two to the third is eight. And now follows a slightly more mathematical procedure.

A simple exponential equation

If we want to solve an exponential equation, it is very convenient if we can adapt the equation to the form $a^{f(x)} = b^{g(x)}$, where a = b. This is because it is obvious that if we have the same bases, the powers will be equal if the exponents are also equal. So we only calculate the equation f(x) = g(x). We can modify the previous example as follows: we subtract the eighth power 2^x = 2³ and then we get the same bases, so we only put the two exponents in the equation and we get the result x = 3. We can make the example more difficult, for example: 2^2x = 8 and after the modification we again get 2^2x = 2³ and calculate 2x = 3, the result is 1.5.

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

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

It looks complicated, but if you know the basic relationships between powers, you know that you can only get one in two ways -- either the base is one and then one on any exponent gives one again, or the exponent is zero -- anything to zero is one again. We'll use the second property, of course, and rewrite the previous equation as follows: 5^{3x − 2} = 5⁰. We have now got the equation into a form that we can already solve (see the previous paragraph). Just compare the exponents. Here comes the equation 3x − 2 = 0. This is already a trivial linear equation.

Logarithming an exponential equation

In case we cannot adapt the equation to a form with the same bases, we can try the logarithm method. Our landlady always said "Kids, if you don't know how to do it, just logarithm it!" and she was right. If you have an exponential equation with different bases, and it is not possible (or not efficient) to adjust them to the same base, logarithm 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 formula for working with logarithms, we get

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

So let's have this example

$$2^x\cdot5^{2x}=3^{x-2}$$

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

$$\log(2^x\cdot5^{2x})=\log(3^{x-2})$$

Now let's use the theorem of logarithms and "multiply" the first parenthesis

$$\log2^x+\log5^{2x}=\log 3^{x-2}$$

Now we use the theorems about logarithms again and "move" the exponent before the logarithm.

$$x\cdot\log2+2x\cdot\log5=(x-2)\log3$$

Multiply the right side of the equation:

$$x\cdot\log2+2x\cdot\log5=x\log3-2\log3$$

Throw the expressions with the unknown onto the left side of the equation:

$$x\cdot\log2+2x\cdot\log5-x\cdot\log3=-2\log3$$

Next, we print x:

$$x(\log2+2\log5-\log3)=-2\log3$$

We isolate x:

$$x=\frac{-2\log3}{\log2+2\log5-\log3}$$

This is already the de facto result. We can reapply the theorems about logarithms and make 2log 5 into $\log 5^2 =\log 25$, but that's really the last thing we can do with the result.

Substitution

We can also solve exponential equations using substitution. Let's show this with an example:

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

Now, for example, we substitute the value a = 7^x for a . Now we modify the original equation by substituting a after 7^x. This is the result of the (no longer exponential) equation:

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

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

We now substitute these roots into the substitution a = 7^x. Thus 2 = 7^x. We logarithm:

$$\log2=x\cdot\log 7$$

and isolate:

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

We don't need to substitute the second root anymore, because it is negative -- substituting it would give us the logarithm with a negative number, which is not possible. And the result of the exponential equation is in the world.

Examples

Calculate the following exponential equation:

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

At first glance, we see that the bases on both sides are not equal. Sad. But at a second glance, we can certainly see the adjustment we can make to get those equal bases. Instead of an eight, we'll simply count 2³, which equals eight. Using the formulas I gave above, specifically the last one, we get:

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

At this point, the bases are already equal and we can calculate the simple linear equation 3x − 4 = 6x + 3:

$$\begin{eqnarray} 3x-4&=&6x+3\\ -3x&=&7\\ x&=&-\frac73 \end{eqnarray}$$

Calculate the following exponential equation:

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

Here, as usual, we see that the bases are not equal. But we probably all suspect that they will somehow convert. Apply the formula for multiplying powers of the same exponent to the left hand side (the second formula in the previous review) and apply the same formula as before to the right hand side, making 100^{x − 1} 10^{2(x − 1)} :

$$\begin{eqnarray} 5^x\cdot2^x&=&100^{x-1}\\ 10^x&=&10^{2(x-1)}\\ 10^x&=&10^{2x-2} \end{eqnarray}$$

And there again we have the same basics and we can calculate the classical linear equation:

$$\begin{eqnarray} x&=&2x-2\\ -x&=&-2\\ x&=&2 \end{eqnarray}$$

Calculate the exponential equation:

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

As always, we have to work our way to the same base first. Here we help ourselves by making an exclamation and extracting from the expression on the left 3^x. Then we add the expression in parentheses and get a simple equation.

$$\begin{eqnarray} 3^x+3^{x+1}&=&108\\ 3^{x}(1+3^1)&=&108\\ 4\cdot3^x&=&108\\ 3^x&=&27 \end{eqnarray}$$

At this point, we've adjusted the left side and it's time to adjust the right side. We can see quite clearly that this is the third power of three:

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

Well, that just leaves the last step - the bases are equal, so we put in the exponents:

$$x=3$$

Calculate the following exponential equation:

$$4^{2x}-6\cdot4^{x}+8=0.$$

In this case, we will try to get an ordinary quadratic equation from the exponential equation. The best way to arrive at it is to use substitution 4^x = a:

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

We now have a standard quadratic equation, so we calculate the discriminant and roots:

$$\begin{eqnarray} D&=&36-4\cdot8\\ a_{1{,}2}&=&\frac{(6\pm2)}{2}\\ a_1&=&4\\ a_2&=&2 \end{eqnarray}$$

We still have to fit these partial results back into the substitution. First result:

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

Second result:

$$\begin{eqnarray} 4^x&=&2\\ 4^x&=&4^{\frac12}\\ x&=&\frac12 \end{eqnarray}$$