Solve Equations with Negative Exponents
We've learnt a lot about solving equations including how to solve one step, two step and three step equations, as well as equations that include fractions. We've also looked at how to group pronumerals (ie. algebraic letters) when they are written on both sides of equation. And then, we've looked at various ways to solve quadratic equations.
We will now look at solving equations with negative exponents (powers). Our goal is to remove the negative exponents, and then use our other equation solving skills to find the solution.
For example, we might want to solve an equation like this:
$x^{-3}-14=13$x−3−14=13
There are two main ways to attack a problem like this.
Method 1 - Turn to positive indices by manipulating fractions.
Remember this index law
$a^{-m}=\frac{1}{a^m}$a−m=1am
So using this law, we change our original equation.
| $x^{-3}-14$x−3−14 | $=$= | $13$13 |
| $\frac{1}{x^3}-14$1x3−14 | $=$= | $13$13 |
From here I employ my normal equation solving skills, isolate the $x$x and then solve.
| $\frac{1}{x^3}-14$1x3−14 | $=$= | $13$13 |
| $\frac{1}{x^3}$1x3 | $=$= | $27$27 |
| $\frac{1}{27}$127 | $=$= | $x^3$x3 |
| $x$x | $=$= | $\frac{1}{3}$13 |
Method 2 - Multiply through the entire equation.
We can remove the negative exponent if we multiply through all terms by it's multiplicative inverse.
For example: $a^{-n}\times a^n=a^{-n+n}=a^0=1$a−n×an=a−n+n=a0=1
So effectively the negative index gets 'cancelled' out, but the by product is we end up with $x$x terms in all the other parts of the equation.
For this example, we need to multiply all the terms by $x^3$x3
| $x^{-3}-14$x−3−14 | $=$= | $13$13 |
| $x^{-3}\times x^3-14x^3$x−3×x3−14x3 | $=$= | $13x^3$13x3 |
| $1-14x^3$1−14x3 | $=$= | $13x^3$13x3 |
| $1$1 | $=$= | $27x^3$27x3 |
| $\frac{1}{27}$127 | $=$= | $x^3$x3 |
| $x$x | $=$= | $\frac{1}{3}$13 |
Sometimes different equations will lend themselves to one particular method of solving over another, but in the end we will always get the same answer regardless of the methods we choose.
Another example
Let's have a look at a slightly harder example $3(x+2)^{-1}+4(x+2)^{-2}$3(x+2)−1+4(x+2)−2
Method 1 - turn to fractions and solve
| $10$10 | $=$= | $3(x+2)^{-1}+4(x+2)^{-2}$3(x+2)−1+4(x+2)−2 |
| $10$10 | $=$= | $\frac{3}{(x+2)}+\frac{4}{(x+2)^2}$3(x+2)+4(x+2)2 |
From here I need a common denominator on the right hand side, and then I will be able to solve.
| $10$10 | $=$= | $\frac{3}{(x+2)}+\frac{4}{(x+2)^2}$3(x+2)+4(x+2)2 |
| $10$10 | $=$= | $\frac{3(x+2)}{(x+2)^2}+\frac{4}{(x+2)^2}$3(x+2)(x+2)2+4(x+2)2 |
| $10(x+2)^2$10(x+2)2 | $=$= | $3(x+2)+4$3(x+2)+4 |
| $10(x^2+4x+4)$10(x2+4x+4) | $=$= | $3x+6+4$3x+6+4 |
| $10x^2+40x+40-3x-10$10x2+40x+40−3x−10 | $=$= | $0$0 |
| $0$0 | $=$= | $10x^2+37x+30$10x2+37x+30 |
Using the quadratic formula I get that $x=\frac{-5}{2}$x=−52 or $\frac{-6}{5}$−65
Method 2 - multiply through
Identifying the largest negative power - we see that it is the term $(x+2)^{-2}$(x+2)−2, so by multiplying through by (x+2)^2 we can remove this term altogether.
REMEMBER - we must multiply ALL TERMS.
| $10$10 | $=$= | $3(x+2)^{-1}+4(x+2)^{-2}$3(x+2)−1+4(x+2)−2 |
| $10(x+2)^2$10(x+2)2 | $=$= | $3(x+2)+4$3(x+2)+4 |
Can you see that by using this process for this equation, we have cut out a number of steps. This is the same process to solve now as above.
This worked nicely because both my terms with the $x$x in it were of the form $(x+2)$(x+2).
Here's an example where the terms have different $x$x constructs.
$-(2x)^{-1}=2(x-2)^{-1}$−(2x)−1=2(x−2)−1
| $-(2x)^{-1}$−(2x)−1 | $=$= | $2(x-2)^{-1}$2(x−2)−1 |
| $\frac{-1}{2x}$−12x | $=$= | $\frac{2}{(x-2)}$2(x−2) |
This time I will multiply through each fraction component separately to remove it.
| $\frac{-1}{2x}\times2x$−12x×2x | $=$= | $\frac{2}{(x-2)}\times2x$2(x−2)×2x |
| $-1$−1 | $=$= | $\frac{4x}{(x-2)}$4x(x−2) |
| $-(x-2)$−(x−2) | $=$= | $4x$4x |
| $-x+2-4x$−x+2−4x | $=$= | $0$0 |
| $-5x+2$−5x+2 | $=$= | $0$0 |
| $x$x | $=$= | $\frac{2}{5}$25 |
An Even Harder Example
$\frac{\frac{4}{9}\left(x^2+2\right)x^{-\frac{4}{7}}-4x^{\frac{10}{7}}}{\left(x^2+2\right)^2}=0$49(x2+2)x−47−4x107(x2+2)2=0
Let's consider the left hand side of the equation, $\frac{\frac{4}{9}\left(x^2+2\right)x^{-\frac{4}{7}}-4x^{\frac{10}{7}}}{\left(x^2+2\right)^2}$49(x2+2)x−47−4x107(x2+2)2, and rewrite the expression without any negative exponents.
Think: In order to rewrite $x^{-\frac{4}{7}}$x−47 with a positive index, we can use the rule $A^{-n}=\frac{1}{A^n}$A−n=1An.
Do: $=\frac{\frac{4\left(x^2+2\right)}{9x^{\frac{4}{7}}}-4x^{\frac{10}{7}}}{\left(x^2+2\right)^2}$=4(x2+2)9x47−4x107(x2+2)2
Think: In order to remove the fraction within the fraction, we can multiply both the numerator and denominator by $9x^{\frac{4}{7}}$9x47.
Do: $=\frac{4\left(x^2+2\right)-4x^{\frac{10}{7}}\times9x^{\frac{4}{7}}}{9x^{\frac{4}{7}}\left(x^2+2\right)^2}$=4(x2+2)−4x107×9x479x47(x2+2)2
Think: The multiplication $x^{\frac{10}{7}}x^{\frac{4}{7}}$x107x47 can be simplified using the rule $A^m\times A^n=A^{m+n}$Am×An=Am+n.
This gives us $x^{\frac{10}{7}}x^{\frac{4}{7}}=x^{\frac{10}{7}+\frac{4}{7}}$x107x47=x107+47$=$=$x^{\frac{14}{7}}=x^2$x147=x2.
Also, $4\times9$4×9 can be simplified at the same time.
Do: $=\frac{4\left(x^2+2\right)-36x^2}{9x^{\frac{4}{7}}\left(x^2+2\right)^2}$=4(x2+2)−36x29x47(x2+2)2
Think: Expand the brackets in the numerator. Leave the denominator in its factored form.
Do: $=\frac{4x^2+8-36x^2}{9x^{\frac{4}{7}}\left(x^2+2\right)^2}$=4x2+8−36x29x47(x2+2)2
Think: Finish off by combining like terms.
Do: $=\frac{-32x^2+8}{9x^{\frac{4}{7}}\left(x^2+2\right)^2}$=−32x2+89x47(x2+2)2
Now that we have simplified the $LHS$LHS, we can solve the original equation, $\frac{\frac{4}{9}\left(x^2+2\right)x^{-\frac{4}{7}}-4x^{\frac{10}{7}}}{\left(x^2+2\right)^2}=0$49(x2+2)x−47−4x107(x2+2)2=0, using our existing equation solving skills.
Think: We want to replace the original $LHS$LHS, $\frac{\frac{4}{9}\left(x^2+2\right)x^{-\frac{4}{7}}-4x^{\frac{10}{7}}}{\left(x^2+2\right)^2}$49(x2+2)x−47−4x107(x2+2)2, with our new $LHS$LHS, $\frac{-32x^2+8}{9x^{\frac{4}{7}}\left(x^2+2\right)^2}$−32x2+89x47(x2+2)2.
Do: $\frac{-32x^2+8}{9x^{\frac{4}{7}}\left(x^2+2\right)^2}=0$−32x2+89x47(x2+2)2=0
Think: Multiply each side by $9x^{\frac{4}{7}}\left(x^2+2\right)^2$9x47(x2+2)2.
This simplifies the $LHS$LHS substantially. The $RHS$RHS remains $0$0.
Do: $-32x^2+8=0$−32x2+8=0
Think: Move the constant term to the other side.
Do: $-32x^2=-8$−32x2=−8
Think: Divide each side by the coefficient of $x$x.
That is, divide each side by -32, which gives the fraction $\frac{-8}{32}=\frac{1}{4}$−832=14.
Do: $x^2=\frac{1}{4}$x2=14
Think: Take the square root of each side.
Remember to take both the positive and negative root.
Do: $x=\pm\frac{1}{2}$x=±12
Mathspace Examples
Question 1
Solve $8x^{-2}-6x^{-1}+1=0$8x−2−6x−1+1=0.
Question 2
Solve $\frac{y^{\frac{5}{7}}-y^{-\frac{2}{7}}}{y}=0$y57−y−27y=0.
Question 3
Consider the expression $\frac{\left(4x+6\right)^{\frac{1}{2}}-2x\left(4x+6\right)^{-\frac{1}{2}}}{4x+6}$(4x+6)12−2x(4x+6)−124x+6.
Rewrite the expression without any negative indices. Give your answer in simplified form using only positive indices.
Hence solve $\frac{\left(4x+6\right)^{\frac{1}{2}}-2x\left(4x+6\right)^{-\frac{1}{2}}}{4x+6}=0$(4x+6)12−2x(4x+6)−124x+6=0.