2.05 Power equations and polynomials
Cubic equations
Similar to solving quadratics we have a few methods to choose between when solving cubic equations:
- Solve using algebraic manipulation - For cubics such as $2x^3+16=0$2x3+16=0. All students should practise this method.
- Factorise - Fully factorising the cubic and then we can then use the null factor law to solve. If $a\times b=0$a×b=0 then either $a=0$a=0 or $b=0$b=0. Students studying Analysis and Approaches should practise this method.
- Technology - Once we have extracted the important information from a question and formed an equation, we could use technology to solve the equation graphically or algebraically. All students should practise this method.
When factorising a cubic recall we can factorise by:
- using the highest common factor
- factor special forms such as the sum and difference of cubes
- identifying a single factor, then using division to establish the remaining quadratic. From here you would employ any of the factorising methods for quadratics.
Power equations
We may have the unknown is in the exponent - exponential equations. Such as:
$3^x=81$3x=81, $5\times2^x=40$5×2x=40 and $7^{5x-2}=20$75x−2=20.
We could use technology to solve these or further tools in algebra called logarithms. However, in many cases algebraic manipulation using our index laws will allow us to solve exponential equations without a calculator. To achieve this we need to write both sides of the equation with the same base, then we can equate the indices.
When both sides of an exponential equation are written with the same base, we can equate the indices:
If $a^x=a^y$ax=ay, then $x=y$x=y.
To write both sides with the same base it is helpful to be familiar with low powers of prime numbers, so you can recognise them and rewrite them in index form. It is also vital to be confident with our index laws.
Worked examples
Example 1
Solve $5^p=125$5p=125, for $p$p.
Think: One side of the equation is a power of $5$5, can we write the other side also as a power of $5$5?
Do:
| $5^p$5p | $=$= | $125$125 | |
| $5^p$5p | $=$= | $5^3$53 | |
| $\therefore$∴ $p$p | $=$= | $3$3 | , by equating indices |
Example 2
Solve $\left(27\right)^{x+1}=\frac{1}{81}$(27)x+1=181, for $x$x.
Think: Can you spot a common base that both sides could be written in? Both $27$27 and $81$81 are powers of $3$3.
Do: Use index laws to write both sides as a single power of three and then equate the indices.
| $\left(27\right)^{x+1}$(27)x+1 | $=$= | $\frac{1}{81}$181 | |
| $\left(3^3\right)^{x+1}$(33)x+1 | $=$= | $\frac{1}{3^4}$134 | |
| $3^{3x+3}$33x+3 | $=$= | $3^{-4}$3−4 | |
| Hence. $3x+3$3x+3 | $=$= | $-4$−4 | , by equating indices |
| $3x$3x | $=$= | $-7$−7 | |
| $\therefore$∴ $x$x | $=$= | $\frac{-7}{3}$−73 |
Example 3
Solve $5\times16^y=40\times\sqrt[3]{32}$5×16y=40×3√32, for $y$y.
Think: We have a power of $2$2 on the left-hand side but it is multiplied by a $5$5. If we first divide both sides by the factor of $5$5, can we then write both sides as powers of $2$2?
Do:
| $5\times16^y$5×16y | $=$= | $40\times\sqrt[3]{32}$40×3√32 |
| $\left(16\right)^y$(16)y | $=$= | $8\times\sqrt[3]{32}$8×3√32 |
We now have an equation with $16$16, $8$8 and $32$32 which can all be written as powers of $2$2. Proceed with index laws and remember $\sqrt[n]{x}=x^{\frac{1}{n}}$n√x=x1n.
| $\left(16\right)^y$(16)y | $=$= | $8\times\sqrt[3]{32}$8×3√32 | |
| $\left(2^4\right)^y$(24)y | $=$= | $2^3\times\left(2^5\right)^{\frac{1}{3}}$23×(25)13 | |
| $2^{4y}$24y | $=$= | $2^{\left(3+\frac{5}{3}\right)}$2(3+53) | |
| $2^{4y}$24y | $=$= | $2^{\frac{14}{3}}$2143 | |
| Hence, $4y$4y | $=$= | $\frac{14}{3}$143 | , by equating indices |
| $\therefore y$∴y | $=$= | $\frac{14}{12}$1412 |
Example 4
Solve $2^{2x}-20\left(2^x\right)+64=0$22x−20(2x)+64=0 for $x$x.
Think: We have index rules for multiplying and dividing powers but not adding. Since we have three terms added together here we will need a different starting approach. Notice the first term can be written as $\left(2^x\right)^2$(2x)2 and the middle term also has a $2^x$2x. In fact, we have a quadratic equation in terms of $2^x$2x. This may be easier to see if we make a substitution.
Do: Let $y=2^x$y=2x, then:
| $2^{2x}-20\left(2^x\right)+64$22x−20(2x)+64 | $=$= | $0$0 |
| $\left(2^x\right)^2-20\left(2^x\right)+64$(2x)2−20(2x)+64 | $=$= | $0$0 |
| Make the substitution $y$y | $=$= | $2^x$2x |
| $y^2-20y+64$y2−20y+64 | $=$= | $0$0 |
| $\left(y-16\right)\left(y-4\right)$(y−16)(y−4) | $=$= | $0$0 |
| $\therefore$∴ $y$y | $=$= | $4$4 or $16$16 |
Since $y=2^x$y=2x, we have $2^x=4$2x=4 or $2^x=16$2x=16, and thus our solutions are $x=2$x=2 or $x=4$x=4.
Example 5
Solve $2x^3-3x^2-32x-15=0$2x3−3x2−32x−15=0 using technology.
Enter the equation into your graphics calculator and you should get the following answers:
Practice questions
Question 1
Solve the equation $\left(2^2\right)^{x+7}=2^3$(22)x+7=23 for $x$x.
Question 2
Solve the equation $25^{x+1}=125^{3x-4}$25x+1=1253x−4.
Question 3
Consider the equation
$\left(2^x\right)^2-9\times2^x+8=0$(2x)2−9×2x+8=0
The equation can be reduced to a quadratic equation by using a certain substitution.
By filling in the gaps, determine the correct substitution that would reduce the equation to a quadratic.
Let $m=\left(\editable{}\right)^{\editable{}}$m=()
Solve the equation for $x$x by using the substitution $m=2^x$m=2x.