# 10.2 Solving Exponential Equations

Exponential equations are often reduced by using radicals—similar to using exponents to solve for radical equations. There is one caveat, though: while odd index roots can be solved for either negative or positive values, even-powered roots can only be taken for even values, but have both positive and negative solutions. This is shown below:

$\begin{array}{l} \text{For odd values of }n,\text{ then }a^n=b\text{ and }a=\sqrt[n]{b} \\ \text{For even values of }n,\text{ then }a^n=b\text{ and }a=\pm \sqrt[n]{b} \end{array}$

Example 10.2.1

Solve for $x$ in the equation $x^5 = 32$.

The solution for this requires that you take the fifth root of both sides.

$\begin{array}{ccc} (x^5)^{\frac{1}{5}}&=&(32)^{\frac{1}{5}} \\ x&=&2 \end{array}$

When taking a positive root, there will be two solutions. For example:

Example 10.2.2

Solve for $x$ in the equation $x^4 = 16$.

The solution for this requires that the fourth root of both sides is taken.

$\begin{array}{rcl} (x^4)^{\frac{1}{4}}&=&(16)^{\frac{1}{4}} \\ \\ x&=&\pm 2 \\ \\ \end{array}$

The answer is $\pm 2$ because $(2)^4=16$ and $(-2)^4=16$.

When encountering more complicated problems that require radical solutions, work the problem so that there is a single power to reduce as the starting point of the solution. This strategy makes for an easier solution.

Example 10.2.3

Solve for $x$ in the equation $2(2x + 4)^2 = 72$.

The first step should be to isolate $(2x+4)^2$, which is done by dividing both sides by 2. This results in $(2x + 4)^2 = 36$.

Once isolated, take the square root of both sides of this equation:

$\begin{array}{rrcrrrr} [(2x&+&4)^2]^{\frac{1}{2}}&=&36^{\frac{1}{2}}&& \\ 2x&+&4&=&\pm 6&& \\ &&2x&=&-4 &\pm &6 \\ &&x&=&-2 &\pm& 3 \\ \\ &&x&=&-5, &1& \end{array}$

Checking these solutions in the original equation indicates that both work.

Example 10.2.4

Solve for $x$ in the equation $(x + 4)^3 + 6 = -119$.

First, isolate $(x + 4)^3$ by subtracting 6 from both sides. This results in $(x + 4)^3 = -125$.

Now, take the cube root of both sides, which leaves:

$\begin{array}{rrrrl} [(x&+&4)^3]^{\frac{1}{3}}&=&[-125]^{\frac{1}{3}} \\ x&+&4&=&-5 \\ &-&4&&-4 \\ \hline &&x&=&-9 \end{array}$

Checking this solution in the original equation indicates that it is a valid solution.

Since you are solving for an odd root, there is only one solution to the cube root of −125. It is only even-powered roots that have both a positive and a negative solution.

# Questions

Solve.

1. $x^2=75$
2. $x^3=-8$
3. $x^2+5=13$
4. $4x^3-2=106$
5. $3x^2+1=73$
6. $(x-4)^2=49$
7. $(x+2)^5=-243$
8. $(5x+1)^4=16$
9. $(2x+5)^3-6=21$
10. $(2x+1)^2+3=21$
11. $(x-1)^{\frac{2}{3}}=16$
12. $(x-1)^{\frac{3}{2}}=8$
13. $(2-x)^{\frac{3}{2}}=27$
14. $(2x+3)^{\frac{4}{3}}=16$
15. $(2x-3)^{\frac{2}{3}}=4$
16. $(3x-2)^{\frac{4}{5}}=16$