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: $$$3\cdot (7^x)^2-(7^x)-4=0 \Rightarrow 3 \cdot 7^{2x}-7^x-4=0$$$