To solve exponential equations, we need to consider the rule of exponents. These rules help us a lot in solving these type of equations.
Contents
- Same Base
- Different Base
- Problem Solving
Same Base
In solving exponential equations, the following theorem is often useful:
If \(a\) is a non-zero constant and \(a^x = a^y,\) then \(x = y.\ _\square\)
We have
\[\begin{align}a^x&=a^y, a\neq 1 \\\frac{a^x}{a^y} &= 1 \\a^{x-y} &= 1 \\x-y &= 0\\x &= y.\ _\square\end{align}\]
Here is how to solve exponential equations:
- Manage the equation using the rule of exponents and some handy theorems in algebra.
- Use the theorem above that we just proved.
Solve \(\displaystyle{ \frac{1}{5^{x-1}} = 125} .\)
Making the bases on both sides equal to 5 gives
\[ \begin{align}\frac{1}{5^{x-1}} &= 125 \\5^{-(x-1)}&=5^3 \\-(x-1) &= 3 \\x &= -2. \ _\square\end{align} \]
Solve \( 4^{x-3} = 0.125 .\)
Converting the bases of both sides to 2 gives
\[ \begin{align}4^{x-3} &= 0.125 \\4^{x-3} &= \frac{125}{1000} \\2^{2x-6} &= \frac{1}{8} \\2^{2x-6} &= 2^{-3} \\2x-6&= -3 \\x&= \frac{3}{2}. \ _\square\end{align} \]
Find the value of \(x\) when \(4^x = 16.\)
We have \(4^x = 16 \Rightarrow 4^x = 4^2.\) Then the theorem "if \(a\) is a non-zero constant and \(a^x = a^y,\) then \(x = y\)" gives \(x = 2.\ _\square\)
If \(8^x = 2\), what is \(x\)?
See AlsoAlgebra - Solving Exponential Equations4.7: Exponential and Logarithmic Equations6.1 Exponential Functions - College Algebra 2e | OpenStaxHow to solve exponential equations with a variable in the exponent. Step 1 is just to...We have
\[\begin{align}
8^x & = 2\\\big(2^3\big)^{x} & = 2\\2^{3x} & = 2^{1}. \end{align}\]Equating the powers, we get
\[\begin{align}
3x & = 1\\x & = \dfrac{1}{3}.\ _\square\end{align}\]
If \(6^x - 1=0,\) what is \(x?\)
We have
\[\begin{align}6^x-1&=0 \\6^x &= 1\\6^x &= 6^0\\x &= 0.\ _\square\end{align}\]
If \((8)(9^x) = 9^x,\) what is \(x?\)
We have
\[\begin{align}(8)(9^x) &= 9^x \\(8)(9^x) - 9^x &= 0 \\(7)9^x &= 0 \\9^x &= 0.\end{align}\]
Then the theorem "if \(a \neq 0\) and \(a^x=0,\) then \(x = \phi\)" gives \(x = \phi.\ _\square\)
Different Base
If the bases are different, there are still techniques for solving these exponential equations. If the bases are powers of a common base, we need only convert one or both bases to the common base and proceed using the "Same Base" case.
Solve \( 4^{3x} =8^{x-1}.\)
We see that while 4 and 8 are different bases, they are both powers of a common base, namely 2. We'll proceed by rewriting 4 and 8 in terms of their common base:
\[ \begin{align}4^{3x} &= 8^{x-1} \\\big(2^2\big)^{3x} &=\big (2^3\big)^{x-1} \\2^{6x} &= 2^{3x-3} \\6x&=3x-3 \\x&= -1. \ _\square\end{align} \]
Solve for \(x:\) \(\dfrac{8^{4x - \sqrt{x}}}{{16}^{2x + \sqrt{x}}} = 2^{2\sqrt{x}}\)
We have
\[\begin{align}
\dfrac{8^{4x - \sqrt{x}}}{{16}^{2x + \sqrt{x}}} & = 2^{2\sqrt{x}}\\ \\\dfrac{{\big(2^3\big)}^{4x - \sqrt{x}}}{{{\big(2^4\big)}}^{2x + \sqrt{x}}} & = 2^{2\sqrt{x}}\\ \\\dfrac{2^{12x - 3\sqrt{x}}}{2^{8x + 4\sqrt{x}}} & = 2^{2\sqrt{x}}\\ \\2^{12x - 3\sqrt{x} - 8x - 4\sqrt{x}} & = 2^{2\sqrt{x}}\\ \\2^{4x - 7\sqrt{x}} & = 2^{2\sqrt{x}}. \end{align}\]Equating the powers gives
\[\begin{align}
4x - 7\sqrt{x} & = 2\sqrt{x}\\4x & = 9\sqrt{x}\\\sqrt{x} & = \dfrac{9}{4}\\x & = \dfrac{81}{16}.\ _\square\end{align}\]
Solve \[\frac{27^{3x-2}}{243} = 81^{3x-6} .\]
Reducing the bases of 27, 81, and 243 on both sides to 3 yields
\[ \begin{align}\frac{27^{3x-2}}{243} &= 81^{3x-6} \\\frac{3^{9x-6}}{3^5} &= 3^{12x-24} \\3^{9x-6-5} &= 3^{12x-24} \\3^{9x-11} &= 3 ^{12x-24} \\9x-11 &= 12x-24 \\3x &= 13 \\x &= \frac{13}{3}. \ _\square\end{align} \]
Unfortunately, it won't always be possible to convert to a common base as we did in the examples above.
For instance, in solving \(5^{x} = 3^{x + 2}\), we note that 5 and 3 are not powers of a nice common base. In this case, we'll need to make use of logarithms.
Solve \(5^{x} = 3^{x + 2}.\)
We have
\[ \begin{align}5^{x} &= 3^{x + 2} \\\log_{10}(5^{x}) &= \log_{10}\big(3^{x + 2}\big) \\x\log_{10}5 &= (x + 2)\log_{10}3 \\x\log_{10}5 &= x\log_{10}3 + 2\log_{10}3 \\x\log_{10}5 - x\log_{10}3 &= 2\log_{10}3 \\x\big(\log_{10}5 - \log_{10}3\big) &= 2\log_{10}3. \end{align} \]
Therefore,
\[x = \frac{2\log_{10}3}{\log_{10}5 - \log_{10}3} \approx 4.3013 .\ _\square\]
Problem Solving
Given \(1728 = 2^a.3^b\), find positive integers \(a\) and \(b\).
We have
\[\begin{align}
1728 & = {12}^{3}\\& = {(4 × 3)}^{3}\\& = 4^3 × 3^3\\& = {(2^2)}^3 × 3^3\\& = 2^6 × 3^3\\\Rightarrow a & = 6, \ b = 3.\ _\square\end{align}\]
Solve \(3^{x^2} = 3^{x}.\)
Since both sides of the equation have the same base, their exponents must also be the same:
\[ \begin{align}3^{x^2} &= 3^{x} \\x^2 &=x \\x^2-x&=0\\x(x-1)&=0\\\Rightarrow x&=0,1. \ _\square\end{align} \]
If \( 2^x \cdot 3^y \cdot 5^z = 45,\) what is value of \( x+y+z?\)
Observe that \(45\) can be factorized as follows:
\[ 45 = 3^2\cdot5=2^0\cdot3^2\cdot5^1.\]
Then \(x=0, y=2, z=1,\) which gives \(x+y+z=0+2+1=3.\) \(_\square\)
If \( x^x \cdot y^y = 108 ,\) what is value of \(x+y ?\)
\(108\) can be factorized as follows:
\[ 108 = 2^2 \cdot 3^3 .\]
This implies that either \(x=2, y=3\) or \(x=3, y=2.\) Therefore the answer is \(x+y=2+3=5. \ _ \square \)
If \( \frac{2^5}{2^3} \cdot 3^0 \cdot 3^1 \cdot 3^2 = 2^x \cdot 3^y ,\) what is value of \(x+y?\)
\( \frac{2^5}{2^3} \) can be rewritten as
\[ \frac{2^5}{2^3} = {2}^{5-3}=2^2. \qquad (1) \]
\( 3^0 \cdot 3^1 \cdot 3^2 \) can be rewritten as
\[ 3^0 \cdot 3^1 \cdot 3^2 = 3^{0+1+2}=3^3. \qquad (2) \]
From \( (1)\) and \( (2),\) \(x\) and \(y\) are
\[ \begin{align}\frac{2^5}{2^3} \cdot 3^0 \cdot 3^1 \cdot 3^2 &= 2^2 \cdot 3^3 \\&= 2^x \cdot 3^y \\\Rightarrow x&= 2,\ y=3.\end{align} \]
Hence, \(x+y= 2+3 = 5. \ _ \square \)
An exponential equation is one in which a variable occurs in the exponent. If both sides of the equation have the same base, then the exponents on both sides are also the same:
\[ a^x=a^y \implies x=y .\]
Here is a list of some rules concerning exponential functions:
\(\quad (1)~~a^m \times a^n = a^{m+n}\)
\(\quad (2)~~a^m \div a^n = a^{m-n}\)
\(\quad (3)~~(a^m)^n = a^{mn}\)
\(\quad (4)~~(ab)^n=a^{n}b^{n}\)
\(\quad (5)~~a^0=b^0\)
\(\quad (6)~~1^m=1^n,\)
where \(a\neq0\) and \(b\neq0.\) Always be cautious of \((5)\) and \((6)\); never forget to check if plugging a zero in an exponent works, or if there are any bases that are equal to 1.
