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:$$a^x=b \Rightarrow x= \log_ab$$

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 $a\neq 1$.

$$2-3^{-x}+3^{x+1}=0$$ 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: $$2-t^{-1}+3 \cdot t=0 \Rightarrow 2 \cdot t - t \cdot t^{-1}+3 \cdot t \cdot t=2t-1+3t^2=0$$ which is an equation of the second degree in the variable $t$.

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

Since it is a second degree equation, we obtain two solutions, and undoing the change for each one we get: $$\displaystyle \left.\begin{array}{rcl} t=\frac{1}{3} & \Rightarrow & 3^x =\frac{1}{3} \\ t=-1 & \Rightarrow & 3^x=-1\end{array}\right\}$$ 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: $$\displaystyle 3^x=\frac{1}{3} \Rightarrow x=\log_3\Big(\frac{1}{3}\Big)=\log_3 1-\log_3 3=0-1=-1$$

$$5^{2x}-2\cdot 5^x-15=0$$ We have used the change of variable $5^x=t$: $$t^2-2t-15=0$$ The second grade equation is solved and we have $$t=5 \mbox{ and } t=-3$$ then $x=\log_5 5 $ and the other solution is not considered since $x=\log_5 (-3)$ makes no sense.

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 $3t^2-t-4=0$ which has solutions $\displaystyle t=\frac{4}{3}$, $t=-1$ and choosing a basis to consider the change $t=7^x$, for example. This way, by substituting in the equation we have: $$3\cdot (7^x)^2-(7^x)-4=0 \Rightarrow 3 \cdot 7^{2x}-7^x-4=0$$