\(2^{x + 5} = 32\)
\begin{align*}
2^{x + 5} & = 32 \\
2^{x + 5} & = 2^{5} \\
\therefore x + 5 & = 5 \\
x & = 0
\end{align*}
2.4 Exponential equations
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2.4 Exponential equations (EMAW)
Exponential equations have the unknown variable in the exponent. Here are some examples:
\begin{align*} {3}^{x + 1} & = 9 \\ {5}^{t} + 3 \times {5}^{t - 1}& = 400 \end{align*}If we can write a single term with the same base on each side of the equation, we can equate the exponents. This is one method to solve exponential equations.
Important: if \(a>0\) and \(a\ne 1\) then:
\begin{align*} {a}^{x}& ={a}^{y} \\ \text{then } x& = y \text{ (same base)} \end{align*}Also notice that if \(a = 1\), then \(x\) and \(y\) can be different.
Worked example 8: Equating exponents
Solve for \(x\): \({3}^{x + 1}=9\).
Change the bases to prime numbers
\[{3}^{x + 1} = {3}^{2}\]The bases are the same so we can equate exponents
\begin{align*} x+1& = 2 \\ \therefore x & = 1 \end{align*}Worked example 9: Equating exponents
Solve for \(t\): \({3}^{t}=1\).
Solve for \(t\)
We know from the exponent identities that \(a^{0}=1\), therefore:
\begin{align*} {3}^{t}& = 1 \\ 3^{t} & = 3^{0} \\ \therefore t &= 0 \end{align*}Worked example 10: Solving equations by taking out a common factor
Solve for \(t\): \({5}^{t} + 3\cdot {5}^{t + 1} = 400\).
Rewrite the expression
\[{5}^{t} + 3\left({5}^{t} \cdot 5\right) = 400\]Take out a common factor
\begin{align*} {5}^{t}\left(1+3\cdot5\right)&=400 \\ {5}^{t}\left(1 + 15\right) &= 400 \end{align*}Simplify
\begin{align*} {5}^{t}\left(16\right) & = 400 \\ {5}^{t} & =25 \end{align*}Change the bases to prime numbers
\[{5}^{t} = {5}^{2}\]The bases are the same so we can equate exponents
\[\therefore t=2\]Worked example 11: Solving equations by factorising a trinomial
Solve for \(x\): \[3^{2x}-80\cdot 3^{x} - 81 = 0\]
Factorise the trinomial
\[(3^{x}-81)(3^{x}+1)=0\]Solve for \(x\)
\(3^x = 81\) or \(3^x = -1\). However \(3^x = -1\) is undefined, so:
\begin{align*} 3^{x}& = 81 \\ {3}^{x} & = 3^{4} \\ x & = 4 \end{align*}Therefore \(x = 4\)
Worked example 12: Solving equations by factorising a trinomial
Solve for \(p\): \[p-13{p}^{\frac{1}{2}} + 36 = 0\]
Rewrite the equation
We notice that \({\left({p}^{\frac{1}{2}}\right)}^{2} = p\) so we can rewrite the equation as:
\[{\left({p}^{\frac{1}{2}}\right)}^{2} - 13{p}^{\frac{1}{2}} + 36 = 0\]Factorise as a trinomial
\[\left({p}^{\frac{1}{2}} - 9\right)\left({p}^{\frac{1}{2}} - 4\right) = 0\]Solve to find both roots
\(\begin{array}{rlcrl} {p}^{\frac{1}{2}} - 9& = 0 & \text{ or } & {p}^{\frac{1}{2}} - 4 & =0 \\ {p}^{\frac{1}{2}} & =9 & & {p}^{\frac{1}{2}} & =4 \\ {\left({p}^{\frac{1}{2}}\right)}^{2} & = {\left(9\right)}^{2} & & {\left({p}^{\frac{1}{2}}\right)}^{2} & = {\left(4\right)}^{2} \\ p & =81 & & p & =16 \end{array}\)Therefore \(p = 81\) or \(p = 16\).
Learners may find Worked Example 13 much easier using the \(k\)-substitution method. You may choose to return to this example once the \(k\)-substitution has been taught.
The solution using \(k\)-substitution is as follows:
\begin{align*} 2^{x}-2^{4-x} &=0 \\ 2^{x}-2^{4} \cdot 2^{-x} & = 0\\ 2^{x}-\dfrac{2^{4}}{2^{x}} & = 0 \\ \text{Let } 2^{x}&=k\\ k-\dfrac{2^{4}}{k} &=0\\ \times k \qquad &k^{2}-16 = 0 \\ (k-4)(k+4)=0\\ k = -4 & \text{ or } \qquad k = 4 \\ 2^{x} \ne -4 & \qquad 2^{x}=4 \\ & \qquad 2^{x}=2^{2}=4 \\ x& = 2 \end{align*}Worked example 13: Solving equations by factorisation
Solve for \(x\): \[2^{x}-2^{4-x} = 0\]
Rewrite the equation
In order to get the equation into a form which we can factorise, we need to rewrite the equation:
\begin{align*} 2^{x}-2^{4-x} &= 0 \\ 2^{x}-2^{4} \cdot 2^{-x} &= 0 \\ 2^{x}-\frac{2^{4}}{2^{x}} &= 0 \end{align*}Now eliminate the fraction by multiplying both sides of the equation by the denominator, \(2^{x}\).
\begin{align*} \left(2^{x}-\frac{2^{4}}{2^{x}}\right) \times 2^{x}&= 0 \times 2^{x} \\ 2^{2x}-16 &= 0 \end{align*}Factorise the equation
Now that we have rearranged the equation, we can see that we are left with a difference of two squares. Therefore:
\begin{align*} 2^{2x}-16 &= 0 \\ (2^x - 4)(2^x+4) &=0 \\ 2^x & = 4 \qquad 2^x \ne -4 \quad \text{(a positive integer with an exponent is always positive)}\\ 2^{x}=2^{2} & =4 \\ x& = 2 \end{align*}Therefore \(x = 2\).
Textbook Exercise 2.3
Solve for the variable:
\(5^{2x + 2} = \dfrac{1}{125}\)
\begin{align*}
5^{2x + 2} & = \dfrac{1}{125} \\
5^{2x + 2} & = \dfrac{1}{5^{3}} \\
5^{2x + 2} & = 5^{-3}\\
\therefore 2x + 2 & = -3 \\
2x & = -5\\
x & = -\dfrac{5}{2}
\end{align*}
\(64^{y + 1} = 16^{2y + 5}\)
\begin{align*}
64^{y + 1} & = 16^{2y + 5} \\
2^{6(y + 1)} & = 2^{4(2y + 5)} \\
2^{6y + 6} & = 2^{8y + 20}\\
\therefore 6y + 6 & = 8y + 20 \\
2y & = -14\\
y & = -7
\end{align*}
\(3^{9x - 2} = 27\)
\begin{align*}
3^{9x - 2} & = 27 \\
3^{9x - 2} & = 3^{3} \\
\therefore 9x - 2 & = 3 \\
9x & = 5\\
x & = \dfrac{5}{9}
\end{align*}
\(\text{25} = \text{5}^ {z -4}\)
\begin{align*} \text{25} & = \text{5}^ {z -4} \\ \text{5}^\text{2} & = \text{5}^ {z -4} \\ \text{2} & = z -4 \\ \text{2} +4 & = z \\ \text{6} & = z \end{align*}
\(- \frac{1}{2} \cdot 6^{\frac{m}{2} + 3} = -18\)
\begin{align*} \left( -2 \right) \left( - \frac{1}{2} 6^{\frac{m}{2} + 3} \right) & = \left( -18 \right) \left( -2 \right) \\ 6^{\frac{m}{2} + 3} & = 36 \\ 6^{\frac{m}{2} + 3} & = 6^{2} \\ \frac{m}{2} + 3 & = 2 \\ \frac{m}{2} & = -1 \\ m & = -2 \end{align*}
\(81^{k + 2} = 27^{k + 4}\)
\begin{align*}
81^{k + 2} & = 27^{k + 4} \\
3^{4(k + 2)} & = 3^{3(k + 4)} \\
\therefore 4k + 8 & = 3k + 12 \\
k & = 4
\end{align*}
\(25^{1 - 2x} - 5^{4} = 0\)
\begin{align*}
25^{1 - 2x} - 5^{4} & = 0 \\
5^{2(1 - 2x)} & = 5^{4} \\
5^{2 - 4x} & = 5^{4} \\
\therefore 2 - 4x & = 4 \\
4x & = -2\\
x & = -\dfrac{1}{2}
\end{align*}
\(27^{x} \times 9^{x - 2} = 1\)
\begin{align*}
27^{x} \times 9^{x - 2} & = 1 \\
3^{3x} \times 3^{2(x - 2)} & = 1 \\
3^{3x + 2x - 4} & = 3^{0} \\
\therefore 5x - 4 & = 0 \\
5x & = 4\\
x & = \frac{4}{5}
\end{align*}
\(2^{t} + 2^{t + 2} = 40\)
\begin{align*}
2^{t} + 2^{t + 2} & = 40 \\
2^{t}(1 + 2^{2}) & = 40\\
2^{t}(5) & = 40 \\
2^{t} & = 8 \\
2^{t} & = 2^{3}\\
\therefore t & = 3
\end{align*}
\((7^x - 49)(3^x - 27) = 0\)
\begin{align*}
(7^x - 49)(3^x - 27) &= 0 \\
(7^x - 7^2)(3^x - 3^3) &= 0 \\
\therefore 7^x - 7^2 = 0 &\text{ or } 3^x - 3^3 = 0 \\
\therefore 7^x = 7^2 &\text{ or } 3^x = 3^3 \\
\therefore x = 2 &\text{ or } x = 3
\end{align*}
\((2 \cdot 2^x - 16)(3^{x+1} - 9) = 0\)
\begin{align*}
(2 \cdot 2^x - 16)(3^{x+1} - 9) &= 0 \\
(2^{x+1} - 2^4)(3^{x+1} - 3^2) &= 0 \\
\therefore 2^{x + 1} - 2^4 = 0 &\text{ or } 3^{x+1} - 3^2 = 0 \\
x + 1 = 4 &\text{ or } x + 1 = 2 \\
\therefore x = 3 &\text{ or } x = 1
\end{align*}
\((10^x - 1)(3^x - 81) = 0\)
\begin{align*}
(10^x - 1)(3^x - 81) &= 0 \\
(10^x - 10^0)(3^x - 3^4)& = 0 \\
\therefore 10^x - 10^0 = 0 &\text{ or } 3^x - 3^4 = 0 \\
\therefore x = 0 &\text{ or } x = 4
\end{align*}
\(2 \times 5^{2 - x} = 5 + 5^{x}\)
\begin{align*}
2 \times 5^{2 - x} & = 5 + 5^{x} \\
2(5^{2})(5^{-x}) & = 5 + 5^{x}\\
\dfrac{2(5^{2})}{5^{x}} - 5 - 5^{x} & = 0 \\
\left(\dfrac{50}{5^{x}}\right) \times 5^{x} -5 \times 5^{x} - 5^{x} \times 5^{x} & = 0 \\
50 - 5(5^{x}) - \left(5^{x}\right)^{2} & = 0 \\
\left(5^{x} - 5\right)\left(5^{x} + 10\right) & = 0 \\
5^{x} - 5 = 0 \text{ or } & 5^{x} + 10 = 0\\
5^{x} = 5 \text{ or } & 5^{x} = -10\\
x = 1 \text{ or } & \text{ undefined}\\
\therefore x & = 1
\end{align*}
\(9^{m} + 3^{3 - 2m} = 28\)
\begin{align*}
9^{m} + 3^{3 - 2m} & = 28 \\
3^{2m} + 3^{3}.3^{-2m} & = 28 \\
3^{2m} + \frac{27}{3^{2m}} - 28 & = 0 \\
\left(3^{2m}\right)^{2} - 28\left(3^{2m}\right) + 27 & = 0 \\
\left(3^{2m} - 27\right)\left(3^{2m} - 1\right) & = 0 \\
3^{2m} - 27 = 0 \text{ or } & 3^{2m} - 1 = 0\\
3^{2m} = 3^{3} \text{ or } & 3^{2m} = 3^{0}\\
2m = 3 \text{ or } & 2m = 0\\
\therefore m & = \dfrac{3}{2} \text{ or } 0
\end{align*}
\(y - 2y^{\frac{1}{2}} + 1 = 0\)
\begin{align*}
y - 2y^{\frac{1}{2}} + 1 & = 0 \\
\left(y^{\frac{1}{2}} - 1\right)\left(y^{\frac{1}{2}} - 1\right) & = 0 \\
y^{\frac{1}{2}} - 1 & = 0\\
y^{\frac{1}{2}} & = 1\\
y^{\frac{1}{2} \times 2} & = 1^{1 \times 2}\\
y & = 1^{2} \\
\therefore y & = 1
\end{align*}
\(4^{x + 3} = \text{0,5}\)
\begin{align*}
4^{x + 3} & = \text{0,5} \\
2^{2x + 6} & = \dfrac{1}{2} \\
2^{2x + 6} & = 2^{-1}\\
\therefore 2x + 6 & = -1 \\
2x & = -7\\
x & = -\dfrac{7}{2}
\end{align*}
\(2^{a} = \text{0,125}\)
\begin{align*}
2^{a} & = \text{0,125} \\
2^{a} & = \dfrac{1}{8} \\
2^{a} & = 2^{-3}\\
\therefore a & = -3
\end{align*}
\(10^{x} = \text{0,001}\)
\begin{align*}
10^{x} & = \text{0,001} \\
10^{x} & = \dfrac{1}{\text{1 000}} \\
10^{x} & = 10^{-3}\\
\therefore x & = -3
\end{align*}
\(2^{x^{2} - 2x - 3} = 1\)
\begin{align*}
2^{x^{2} - 2x - 3} & = 1 \\
2^{x^{2} - 2x - 3} & = 2^{0} \\
\therefore x^{2} - 2x - 3 & = 0 \\
(x - 3)(x + 1) & = 0\\
\therefore x & = 3 \text{ or } -1
\end{align*}
\(\dfrac{8^x - 1}{2^x - 1} = 8.2^x + 9\)
\begin{align*}
\frac{8^x - 1}{2^x - 1} &= 8.2^x + 9 \\
\frac{2^{3x} - 1}{2^x - 1} &= 8.2^x + 9 \\
\frac{(2^{x} - 1)(2^{2x} + 2^x + 1)}{2^x - 1} &= 8.2^x + 9 \\
2^{2x} + 2^x + 1 &= 8.2^x + 9 \\
2^{2x} + 2^x &= 8.2^x + 8 \\
2^x.2^x + 2^x &= 8(2^x + 1) \\
2^x(2^x + 1) &= 8(2^x + 1) \\
2^x &= 2^3 \\
\therefore x &= 3
\end{align*}
\(\dfrac{27^x - 1}{9^x + 3^x + 1} = -\dfrac{8}{9}\)
\begin{align*}
\frac{27^x - 1}{9^x + 3^x + 1} &= -\frac{8}{9} \\
\frac{3^{3x} - 1}{9^x + 3^x + 1} &= -\frac{8}{9} \\
\frac{(3^x - 1)(9^x + 3^x + 1)}{9^x + 3^x + 1} &= -\frac{8}{9} \\
3^x - 1 &= -\frac{8}{9} \\
3^x &= \frac{1}{9} \\
3^x &= \frac{1}{3^2} \\
3^x &= 3^{-2} \\
\therefore x &= -2
\end{align*}
The growth of algae can be modelled by the function \(f(t) = 2^{t}\). Find the value of \(t\) such that \(f(t) = 128\).
\begin{align*}
f(t) = 2^{t} & = 128 \\
2^{t} & = 2^{7} \\
\therefore t & = 7
\end{align*}
Use trial and error to find the value of \(x\) correct to 2 decimal places
\(2^x = 7\)
\begin{align*}
2^2 = 4 &\text{ and } 2^3= 8 \\
\text{so } 2 < &x < 3 \text{ but closer to }3 \\
\text{Test} \\
2^{\text{2,9}} &= \text{7,464} \\
2^{\text{2,8}} &= \text{6,964} \\
2^{\text{2,81}} &= \text{7,01} \\
2^{\text{2,805}} &= \text{6,989} \\
2^{\text{2,809}} &= \text{7,007} \\
\therefore x &\approx \text{2,81}
\end{align*}
Use trial and error to find the value of \(x\) correct to 2 decimal places
\(5^x = 11\)
\begin{align*}
5^1 = 5 &\text{ and } 5^2 = 25 \\
\text{so } 1 < &x < 2 \\
\text{Test} \\
5^{\text{1,5}} &= \text{11,180} \\
5^{\text{1,4}} &= \text{9,51} \\
5^{\text{1,45}} &= \text{10,31} \\
5^{\text{1,49}} &= \text{11,001} \\
\therefore x &\approx \text{1,49}
\end{align*}
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