# Basic Exponential Equations

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Objectives
• Solve for variable exponents in an equation using only what we know about integers and fractions
• Learn the "same base" trick for solving certain equations that involve variable exponents, and know when the trick can and can't be used
• Understand that when the "same base" trick cannot be used, that there is necessarily no rational solution (solution requires forthcoming logarithm techniques)
• Solve special kinds of exponential equations that are quadratic equations in disguise
Lesson Description

Exponential equations are equations in which the unknown variable is an exponent. There are two ways to solve these: either using integer and fraction knowledge or using logarithms. While we will study logarithms very soon, this lesson shows us how to solve exponential equations using only what we know about integers and fractions. We'll also understand when we can use this technique, and when we instead need to rely on logarithms.

Practice Problems

Practice problems and worksheet coming soon!

## Solving for Variable Exponents

Now that we're better acquainted with the concept » of variable exponents, it's time to start solving for them in equations.Equations that have $x$ as an exponent to solve for come in two flavors: the first requires only arithmetic and logic, which we will study in this lesson. The second requires the use of logarithms » which are on deck in the Mr. Math curriculum.

## Simple Exponent Logic

To solve exponential equations using solely logic and arithmetic, the equation must contain bases that are the same or related. Recall that bases are the numbers that are raised to powers.The logic that we'll use is simply this: if the same base is raised to two different powers, but the two expressions are equal, then the exponents must be also equal.
Theorem: Same Base EquationsIf an both sides of an equation consist only of the same number raised to an exponent, then the exponents are equal.Symbolically, if$$b^x = b^y$$then it follows that$$x=y$$where $b$ is any real number.

Example 1Solve for $x$.$$5^x = 5^3$$$\blacktriangleright$ In this first, one-step example, let's better understand the previous logic statement.If the same base, $5$, is raised to two different powers, $x$ on one side and $3$ on the other, but the equation tells us that the expressions are equal, then the exponents must also be equal.In this case, we know $x$ must be $3$.

Hopefully Example 1 clarified how the "logic" works. More arithmetic will be involved if the bases of each expression are related instead of equal.

Example 2Solve$$3^{6x+1} = 81$$$\blacktriangleright$ While we will use a similar approach, we cannot one-step this problem. First we have to re-write it so that $3$ and $81$ are the same.Since $81$ is $3^4$, we can set up the following two steps;$$3^{6x+1} = 3^4$$$$\longrightarrow 6x +1 = 4$$In words, we're applying that logic once more. If $3$ raised to a certain power is equal to $3$ raised to a different power, then the powers must be equal.Finally,$$6x+1 = 4$$$$6x = 3$$$$x = \frac{1}{2}$$

You Should Know When it comes to solving for variable exponents, you'll either be able to solve it this way logarithm-free, or you will need logarithms. This method is preferable if it is applicable because it's quick and digestible, but technically you can use logarithms to solve any variable exponent equation, including these ones (it's just more work to do so).

## Typical Test Questions

To bring these problems up to the "full difficulty" of typical test expectations, we need to look at problems that require changes to both sides of the equation.

Example 3Solve$$4^{7-x} = 32^{2x-2}$$$\blacktriangleright$ Not only can we not set these powers equal right away, but we can't directly change either base to the other. However, $4$ and $32$ are related, so break down each into their common base: $2$.Since $4=2^2$ and $32=2^5$, we can re-write:$$\left(2^2\right)^{7-x} = \left(2^5\right)^{2x-2}$$Recall the rules of exponents, and what they tell us to do for powers raised to powers - we multiply the powers together. Now we have$$2^{14-2x} = 2^{10x-10}$$Now we can set the powers equal using the logic.$$14-2x = 10x-10$$$$24 = 12x$$$$x=2$$

## Mr. Math Makes It Mean

Secret QuadraticsBecause we're dealing with exponent rules, it's super easy to make quadratic equations in disguise using exponential terms. These questions often appear on tests as difficult or bonus questions and throw students off because they are so odd and specific. Let's dive into one.

Example 4Solve$$3^{2x} - 12 \left( 3^x \right) + 30 = 3$$$\blacktriangleright$ Because $3^{2x}$ can be written as $3^{x}$ squared, we can rewrite this and use the reverse FOIL method of solving the resulting quadratic.$$\left(3^x\right)^2 - 12 \left( 3^x \right) + 30 = 3$$Move the $3$ to the left side and factor.$$\left(3^x\right)^2 - 12 \left( 3^x \right) + 27 = 0$$$$\left(3^x - 3\right) \left(3^x - 9 \right) = 0$$This factored equation will be satisfied when either factor is equal to zero (the zero-product property). When $3^x - 3 = 0$, $x$ must be $1$. When $3^x - 9 = 0$, $x$ must be $2$. Therefore there are two answers to this equation, and $x$ can be either $1$ or $2$.

## Unusual Suspects

It isn't common but some teachers will really emphasize understanding when you can or cannot avoid using logarithms.The key here is to answer a yes/no question. Either the bases can or cannot be re-written to a common base.

Example 5Determine which of the following equations can and cannot be solved using only arithmetic and logic. Do not solve any of these equations.\begin{align} \mathrm{a)} \;\;\;\; & 49^{2x} = 243^{3x} \\ \mathrm{b)} \;\;\;\; & 6^{2x+1} = 72^{3x-2} \\ \mathrm{c)} \;\;\;\; & 6561^{10x -7} = 729^{4x + 1} \end{align}$\blacktriangleright$ The first equation can be written entirely in base $7$, since $49=7^2$ and $243=7^3$. This is solvable using only what we have learned in this lesson.The second equation is only solvable with logarithms, because $6$ and $72$ cannot be made into the same base.The third equation is doable with what we know, since both $6561$ and $729$ can be written base $3$ (or even base $9$).

## Put It To The Test

Example 6Solve$$9^x = 243$$
Show solution
$\blacktriangleright$ This two-step problem requires us to do what we usually do - rewrite each side as the same base. In that case, the right choice is base $3$.Since $9 = 3^2$ and $243 = 3^5$, we really have:$$\left(3^2\right)^x = 3^5$$$$\longrightarrow 3^{2x} = 3^5$$$$\longrightarrow 2x = 5$$$$x = \frac{5}{2}$$

Example 7Solve$$125^x = 25^{x/2 + 5}$$
Show solution
$\blacktriangleright$ Both sides can be written in base $5$:$$125 = 5^3$$$$25 = 5^2$$Therefore, we can rewrite the problem as$$\left( 5^3 \right)^x = \left(5^2 \right)^{x/2 + 5}$$$$\longrightarrow 5^{3x} = 5^{x + 10}$$$$\longrightarrow 3x = x + 10$$$$2x = 10$$$$x = 5$$

Example 8Solve$$4^{2x+1} = 8^{3x+4}$$
Show solution
$\blacktriangleright$ Let's rewrite each side with the same base of $2$, since $4=2^2$, and $8=2^3$.$$\left( 2^2 \right)^{2x+1} = \left( 2^3 \right)^{3x+4}$$$$\longrightarrow 2^{4x+2} = 2^{9x+12}$$$$\longrightarrow 4x+2 = 9x+12$$$$5x = 10$$$$x=2$$

Example 9Solve$$2^{2x} -20\left(2^x\right) + 64 = 0$$
Show solution
$\blacktriangleright$ This is one of those quadratic equations in disguise. Since $2^{2x}$ can be written as $(2^x)^2$, we can follow a quadratic factoring pattern:$$\left(2^{x}\right)^2 -20\left(3^x\right) + 64 = 0$$$$\longrightarrow \left( 2^x - 4\right) \left( 2^x - 16 \right) = 0$$To make this equation true, one of two things must be true. Either $2^x = 4$ which implies that $x$ is $2$, or $2^x = 16$, which implies that $x = 4$. Therefore there are two answers to this problem: $x$ can be either $2$ or $4$.

Lesson Takeaways
• Learn and remember the logic that allows basic exponential equations to be solved using only arithmetic
• Know when an equation can and cannot be solved using arithmetic and logic only, as opposed to requiring logarithms
• Be familiar with the ways in which teachers expect you to solve these types of equations on tests
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