# Solving an exponential equation by reducing to an equal base

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, in both sides of the exponential equation to be solved, there are powers of the same basis: in this case, one proceeds bylimiting both sides of the equation to the same basis and modifying the exponents until an equality of basis is achieved, which allows us to equal the exponents.

$$2^{x+1}+2^x+2^{x-1}=28$$$On the left part of the equality we extract a common factor of $$2^x$$, and on the right part we leave it as it is. And so, $$2^{x+1}+2^x+2^{x-1}=28 \mbox { is the same as }2\cdot2^x+2^x+\displaystyle \frac{2^x}{2}=28 \Rightarrow 2^x\Big(2+1+\frac{1}{2}\Big)=28$$$

Now we perform the operations and isolate $$2^x$$:

$$2^x\Big(2+1+\frac{1}{2}\Big)=28 \Rightarrow 2^x\cdot \frac{7}{2}=28 \Rightarrow 2^x=\frac{28\cdot 2}{7}=8=2^3$$$We already have the same base on both sides of the equality, so that we can equal the exponents. In this way: $$2^x=2^3\Rightarrow x=3$$$

which is the solution we were looking for.

$$4^x=8^{2x-3} \Rightarrow 2^{2x}=2^{3(2x-3)} \Rightarrow 2x=3(2x-3) \Rightarrow 4x=9 \rightarrow \displaystyle x=\frac{9}{4}$$$To construct a solvable equation of this form, it is possible to proceed in the inverse order. We choose a basis, for example $$3$$. We invent an equation of the first degree: $$5\Big(x+\frac{4}{7}\Big)=21$$$

and we write it as an exponent of such a basis:

$$\displaystyle 3^{5\Big(x+\frac{4}{7}\Big)}=3^21$$$Now we can try to write the same in a different form so that a few calculations have to be done before coming up with the same basis. For example, a possibility would be: $$\displaystyle 3^{5\Big(x+\frac{4}{7}\Big)}=3^{5x}\cdot 3^\frac{5\cdot 4}{7}=243^x\sqrt[7]{3^{20}} \\ 3^{21}=3^{17+4}=3^{5\cdot 3+2}\cdot 3^4=9\cdot (3^5)^4\cdot \frac{3^8}{3^4}$$$

which, when equaling them, will give us the following equation:

$$\displaystyle 243^x \cdot \sqrt[7]{3^{20}}=9 \cdot (3^5)^3\cdot \frac{3^8}{3^5}$$\$

There are many possible combinations for equations with the same solution.