Solving an exponential equation by variable change

An exponential equation is one in which the unknown variable or variables is/are in the exponent of a power. The exponential equations use basic knowledge of the exponential and logarithmic functions. As such, we will revise them.

To solve them the following properties are used:

$a^{0} = 1$ for any $a$ .
Two potencies with the samepositivebasis and different from the unit are equal, if, and only if, its exponents are equal. That is to say: $2^{a} = 2^{b} \Leftrightarrow a = b$
For any $a \neq 0$ and $a \neq 1$ we have: $a^{x} = b \Rightarrow x = \log_{a} b$

When it comes to solving an exponential equation it can have different forms and, because of that, there are different methods and transformations.

When the equation is of the type $f (a^{x}) = 0$ ,the change of variable $t = a^{x}$ is used and the equation of the first or second order that appearsis then solved. Whenever we want to appl, one must ensure that $a \neq 0$ and $a \neq 1$ .

Example

$2 - 3^{- x} + 3^{x + 1} = 0$ is of this type since $f (3^{x}) = 2 - 3^{- x} + 3^{x + 1} = 0$

We have used the change of variable $t = 3^{x}$ .

Then: $2 - (3^{x})^{- 1} + 3 \cdot 3^{x} = 2 - t^{- 1} + 3 \cdot t = 0$

Since we are sure that $t$ is not zero (because any $x$ such that $3^{x}$ is zero does not exist) there is no problem where $t$ appears in the denominator. Multiplying the expression by $t$ we obtain: $2 - t^{- 1} + 3 \cdot t = 0 \Rightarrow 2 \cdot t - t \cdot t^{- 1} + 3 \cdot t \cdot t = 2 t - 1 + 3 t^{2} = 0$ which is an equation of the second degree in the variable $t$ .

We solve it: $t = \frac{- 2 \pm \sqrt{2^{2} - 4 \cdot 3 \cdot (- 1)}}{2 \cdot 3} = - 2 \pm \frac{\sqrt{4 + 12}}{6} = \frac{- 2 \pm \sqrt{16}}{6} =$ $= \frac{- 2 \pm 4}{6} = {\begin{cases} t = \frac{2}{6} = \frac{1}{3} \\ t = \frac{- 6}{6} = - 1 \end{cases}$

Since it is a second degree equation, we obtain two solutions, and undoing the change for each one we get: $\begin{array}{rcl} t = \frac{1}{3} & \Rightarrow & 3^{x} = \frac{1}{3} \\ t = - 1 & \Rightarrow & 3^{x} = - 1 \end{array}}$ but $3^{x}$ can never be a negative value so that asolution does not exist for $t = - 1$ . Now we only have one solution which is: $3^{x} = \frac{1}{3} \Rightarrow x = \log_{3} (\frac{1}{3}) = \log_{3} 1 - \log_{3} 3 = 0 - 1 = - 1$

Example

$5^{2 x} - 2 \cdot 5^{x} - 15 = 0$ We have used the change of variable $5^{x} = t$ : $t^{2} - 2 t - 15 = 0$ The second grade equation is solved and we have $t = 5 and t = - 3$ then $x = \log_{5} 5$ and the other solution is not considered since $x = \log_{5} (- 3)$ makes no sense.

Example

Let's imagine that we want to construct an equation that is solved in this way. A way of proceeding involves taking an equation of second degree $3 t^{2} - t - 4 = 0$ which has solutions $t = \frac{4}{3}$ , $t = - 1$ and choosing a basis to consider the change $t = 7^{x}$ , for example. This way, by substituting in the equation we have: $3 \cdot (7^{x})^{2} - (7^{x}) - 4 = 0 \Rightarrow 3 \cdot 7^{2 x} - 7^{x} - 4 = 0$