Chapter 10: Quadratics

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:

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

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

<a class=”internal” href=”/intermediatealgebraberg/back-matter/answer-key-10-2/”>Answer Key 10.2

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