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Exponential Equations

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 \(x = 2\).

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Exercise 2.3

Solve for the variable:

\(2^{x + 5} = 32\)

\begin{align*} 2^{x + 5} & = 32 \\ 2^{x + 5} & = 2^{5} \\ \therefore x + 5 & = 5 \\ x & = 0 \end{align*}

\(5^{2x + 2} = \dfrac{1}{125}\)

\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*}