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
The solution for this requires that you take the fifth root of both sides.
[latex]\begin{array}{ccc} (x^5)^{\frac{1}{5}}&=&(32)^{\frac{1}{5}} \\ x&=&2 \end{array}[/latex]
When taking a positive root, there will be two solutions. For example:
Example 10.2.2
Solve for [latex]x[/latex] in the equation [latex]x^4 = 16[/latex].
The solution for this requires that the fourth root of both sides is taken.
[latex]\begin{array}{rcl} (x^4)^{\frac{1}{4}}&=&(16)^{\frac{1}{4}} \\ \\ x&=&\pm 2 \\ \\ \end{array}[/latex]
The answer is [latex]\pm 2[/latex] because [latex](2)^4=16[/latex] and [latex](-2)^4=16[/latex].
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 [latex]x[/latex] in the equation [latex]2(2x + 4)^2 = 72[/latex].
The first step should be to isolate [latex](2x+4)^2[/latex], which is done by dividing both sides by 2. This results in [latex](2x + 4)^2 = 36[/latex].
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 [latex]x[/latex] in the equation [latex](x + 4)^3 + 6 = -119[/latex].
First, isolate [latex](x + 4)^3[/latex] by subtracting 6 from both sides. This results in [latex](x + 4)^3 = -125[/latex].
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.
- [latex]x^2=75[/latex]
- [latex]x^3=-8[/latex]
- [latex]x^2+5=13[/latex]
- [latex]4x^3-2=106[/latex]
- [latex]3x^2+1=73[/latex]
- [latex](x-4)^2=49[/latex]
- [latex](x+2)^5=-243[/latex]
- [latex](5x+1)^4=16[/latex]
- [latex](2x+5)^3-6=21[/latex]
- [latex](2x+1)^2+3=21[/latex]
- [latex](x-1)^{\frac{2}{3}}=16[/latex]
- [latex](x-1)^{\frac{3}{2}}=8[/latex]
- [latex](2-x)^{\frac{3}{2}}=27[/latex]
- [latex](2x+3)^{\frac{4}{3}}=16[/latex]
- [latex](2x-3)^{\frac{2}{3}}=4[/latex]
- [latex](3x-2)^{\frac{4}{5}}=16[/latex]
