IB · Unit 1 · 1A Exponents
Rewrite both equations with the same base. That gives a linear system in the exponents.
Product and quotient
Same base, multiply: add the exponents. Same base, divide: subtract the exponents.
\(a^x \cdot a^y = a^{x+y}\)
\(\dfrac{a^x}{a^y} = a^{x-y}\)
Equation 1. \(2^x \cdot 2^y = 8\)
Equation 2. \(\dfrac{2^x}{2^y} = 2\)
Equation 1. Product: add the exponents. Change 8 to a power of 2.
\(2^x \cdot 2^y = 2^3\)
\(2^{x+y} = 2^3\)
Bases are equal, so x + y = 3.
Equation 2. Quotient: subtract the exponents. 2 is 2 to the first power.
\(\dfrac{2^x}{2^y} = 2^1\)
\(2^{x-y} = 2^1\)
Bases are equal, so x − y = 1.
Add: 2x = 4, so x = 2 and y = 1.
When the right side is 1
Students get stuck on \(2^x \cdot 8^y = 1\) because they forget that 1 can be rewritten as \(2^0\).
\(2^x \cdot 8^y = 1\)
\(2^x \cdot (2^3)^y = 2^0\)
\(2^x \cdot 2^{3y} = 2^0\)
\(2^{x+3y} = 2^0\)
\(x + 3y = 0\)
The bases already match. Product: add the exponents. Quotient: subtract the exponents.
Equation 1.\(2^x \cdot 2^y = 8\)
\(2^x \cdot 2^y = 2^3\)
\(2^{x+y} = 2^3\)
\(x + y = 3\)
Equation 2.\(\dfrac{2^x}{2^y} = 32\)
\(\dfrac{2^x}{2^y} = 2^5\)
\(2^{x-y} = 2^5\)
\(x - y = 5\)
Add Equation 1 and Equation 2.\[\begin{array}{r@{\;}r@{}c@{}l}& x + y &=& 3 \\+& x - y &=& 5\end{array}\]
\(2x = 8\)
\(x = 4\)
Substitute x = 4 into Equation 1.\(4 + y = 3\)
\(y = -1\)
Answer: x = 4, y = −1
Equation 1.\(3^x \cdot 3^y = 81\)
\(3^x \cdot 3^y = 3^4\)
\(3^{x+y} = 3^4\)
\(x + y = 4\)
Equation 2.\(\dfrac{3^x}{3^y} = 9\)
\(\dfrac{3^x}{3^y} = 3^2\)
\(3^{x-y} = 3^2\)
\(x - y = 2\)
Add Equation 1 and Equation 2.\[\begin{array}{r@{\;}r@{}c@{}l}& x + y &=& 4 \\+& x - y &=& 2\end{array}\]
\(2x = 6\)
\(x = 3\)
Substitute x = 3 into Equation 1.\(3 + y = 4\)
\(y = 1\)
Answer: x = 3, y = 1
Equation 1.\(5^{x+y} = 25\)
\(5^{x+y} = 5^2\)
\(x + y = 2\)
Equation 2.\(5^{x-y} = 5\)
\(5^{x-y} = 5^1\)
\(x - y = 1\)
Add Equation 1 and Equation 2.\[\begin{array}{r@{\;}r@{}c@{}l}& x + y &=& 2 \\+& x - y &=& 1\end{array}\]
\(2x = 3\)
\(x = \dfrac{3}{2}\)
Substitute into Equation 1.\(\dfrac{3}{2} + y = 2\)
\(y = \dfrac{1}{2}\)
Answer: \(x = \dfrac{3}{2},\ y = \dfrac{1}{2}\)
Equation 1.\(2^x \cdot 2^y = 16\)
\(2^x \cdot 2^y = 2^4\)
\(2^{x+y} = 2^4\)
\(x + y = 4\)
Equation 2.\(\dfrac{2^x}{2^y} = 4\)
\(\dfrac{2^x}{2^y} = 2^2\)
\(2^{x-y} = 2^2\)
\(x - y = 2\)
Add Equation 1 and Equation 2.\[\begin{array}{r@{\;}r@{}c@{}l}& x + y &=& 4 \\+& x - y &=& 2\end{array}\]
\(2x = 6\)
\(x = 3\)
Substitute x = 3 into Equation 1.\(3 + y = 4\)
\(y = 1\)
Answer: x = 3, y = 1
Equation 1.\(7^{x+y} = 7\)
\(7^{x+y} = 7^1\)
\(x + y = 1\)
Equation 2.\(7^{x-y} = 343\)
\(7^{x-y} = 7^3\)
\(x - y = 3\)
Add Equation 1 and Equation 2.\[\begin{array}{r@{\;}r@{}c@{}l}& x + y &=& 1 \\+& x - y &=& 3\end{array}\]
\(2x = 4\)
\(x = 2\)
Substitute x = 2 into Equation 1.\(2 + y = 1\)
\(y = -1\)
Answer: x = 2, y = −1
Equation 1.\(4^x \cdot 4^y = 64\)
\(4^x \cdot 4^y = 4^3\)
\(4^{x+y} = 4^3\)
\(x + y = 3\)
Equation 2.\(4^{x-y} = 4\)
\(4^{x-y} = 4^1\)
\(x - y = 1\)
Add Equation 1 and Equation 2.\[\begin{array}{r@{\;}r@{}c@{}l}& x + y &=& 3 \\+& x - y &=& 1\end{array}\]
\(2x = 4\)
\(x = 2\)
Substitute x = 2 into Equation 1.\(2 + y = 3\)
\(y = 1\)
Answer: x = 2, y = 1
Rewrite 4, 8, or 9 as a power of 2 or 3 first. To eliminate, multiply one equation by −1, then add.
Equation 1.\(4^x \cdot 2^y = 32\)
\((2^2)^x \cdot 2^y = 2^5\)
\(2^{2x} \cdot 2^y = 2^5\)
\(2^{2x+y} = 2^5\)
\(2x + y = 5\)
Equation 2.\(2^x \cdot 4^y = 16\)
\(2^x \cdot (2^2)^y = 2^4\)
\(2^x \cdot 2^{2y} = 2^4\)
\(2^{x+2y} = 2^4\)
\(x + 2y = 4\)
Multiply Equation 2 by 2.\(2(x + 2y) = 8\)
\(2x + 4y = 8\)
Multiply Equation 1 by −1.\(-(2x + y) = -5\)
\(-2x - y = -5\)
Add.\[\begin{array}{r@{\;}r@{}c@{}l}& 2x + 4y &=& 8 \\+& -2x - y &=& -5\end{array}\]
\(3y = 3\)
\(y = 1\)
Substitute y = 1 into Equation 1.\(2x + 1 = 5\)
\(2x = 4\)
\(x = 2\)
Answer: x = 2, y = 1
Equation 1.\(9^x \cdot 3^y = 243\)
\((3^2)^x \cdot 3^y = 3^5\)
\(3^{2x} \cdot 3^y = 3^5\)
\(3^{2x+y} = 3^5\)
\(2x + y = 5\)
Equation 2.\(3^x \cdot 9^y = 81\)
\(3^x \cdot (3^2)^y = 3^4\)
\(3^x \cdot 3^{2y} = 3^4\)
\(3^{x+2y} = 3^4\)
\(x + 2y = 4\)
Multiply Equation 2 by 2.\(2(x + 2y) = 8\)
\(2x + 4y = 8\)
Multiply Equation 1 by −1.\(-(2x + y) = -5\)
\(-2x - y = -5\)
Add.\[\begin{array}{r@{\;}r@{}c@{}l}& 2x + 4y &=& 8 \\+& -2x - y &=& -5\end{array}\]
\(3y = 3\)
\(y = 1\)
Substitute y = 1 into Equation 1.\(2x + 1 = 5\)
\(2x = 4\)
\(x = 2\)
Answer: x = 2, y = 1
Equation 1.\(4^x \cdot 8^y = 256\)
\((2^2)^x \cdot (2^3)^y = 2^8\)
\(2^{2x} \cdot 2^{3y} = 2^8\)
\(2^{2x+3y} = 2^8\)
\(2x + 3y = 8\)
Equation 2.\(8^x \cdot 4^y = 128\)
\((2^3)^x \cdot (2^2)^y = 2^7\)
\(2^{3x} \cdot 2^{2y} = 2^7\)
\(2^{3x+2y} = 2^7\)
\(3x + 2y = 7\)
Multiply Equation 1 by 2.\(2(2x + 3y) = 16\)
\(4x + 6y = 16\)
Multiply Equation 2 by 3.\(3(3x + 2y) = 21\)
\(9x + 6y = 21\)
Multiply 4x + 6y = 16 by −1.\(-(4x + 6y) = -16\)
\(-4x - 6y = -16\)
Add.\[\begin{array}{r@{\;}r@{}c@{}l}& 9x + 6y &=& 21 \\+& -4x - 6y &=& -16\end{array}\]
\(5x = 5\)
\(x = 1\)
Substitute x = 1 into Equation 1.\(2(1) + 3y = 8\)
\(2 + 3y = 8\)
\(3y = 6\)
\(y = 2\)
Answer: x = 1, y = 2
Equation 1.\(\dfrac{9^x}{3^y} = 27\)
\(\dfrac{(3^2)^x}{3^y} = 3^3\)
\(\dfrac{3^{2x}}{3^y} = 3^3\)
\(3^{2x-y} = 3^3\)
\(2x - y = 3\)
Equation 2.\(3^x \cdot 9^y = 81\)
\(3^x \cdot (3^2)^y = 3^4\)
\(3^x \cdot 3^{2y} = 3^4\)
\(3^{x+2y} = 3^4\)
\(x + 2y = 4\)
Multiply Equation 1 by 2.\(2(2x - y) = 6\)
\(4x - 2y = 6\)
Add.\[\begin{array}{r@{\;}r@{}c@{}l}& 4x - 2y &=& 6 \\+& x + 2y &=& 4\end{array}\]
\(5x = 10\)
\(x = 2\)
Substitute x = 2 into Equation 1.\(2(2) - y = 3\)
\(4 - y = 3\)
\(y = 1\)
Answer: x = 2, y = 1
Equation 1.\(\dfrac{8^x}{2^y} = 4\)
\(\dfrac{(2^3)^x}{2^y} = 2^2\)
\(\dfrac{2^{3x}}{2^y} = 2^2\)
\(2^{3x-y} = 2^2\)
\(3x - y = 2\)
Equation 2.\(2^x \cdot 4^y = 8\)
\(2^x \cdot (2^2)^y = 2^3\)
\(2^x \cdot 2^{2y} = 2^3\)
\(2^{x+2y} = 2^3\)
\(x + 2y = 3\)
Multiply Equation 1 by 2.\(2(3x - y) = 4\)
\(6x - 2y = 4\)
Add.\[\begin{array}{r@{\;}r@{}c@{}l}& 6x - 2y &=& 4 \\+& x + 2y &=& 3\end{array}\]
\(7x = 7\)
\(x = 1\)
Substitute x = 1 into Equation 1.\(3(1) - y = 2\)
\(3 - y = 2\)
\(y = 1\)
Answer: x = 1, y = 1
Equation 1.\(4^x \cdot 2^y = 8\)
\((2^2)^x \cdot 2^y = 2^3\)
\(2^{2x} \cdot 2^y = 2^3\)
\(2^{2x+y} = 2^3\)
\(2x + y = 3\)
Equation 2.\(\dfrac{2^x}{4^y} = \dfrac{1}{2}\)
\(\dfrac{2^x}{(2^2)^y} = 2^{-1}\)
\(\dfrac{2^x}{2^{2y}} = 2^{-1}\)
\(2^{x-2y} = 2^{-1}\)
\(x - 2y = -1\)
Multiply Equation 1 by 2.\(2(2x + y) = 6\)
\(4x + 2y = 6\)
Add.\[\begin{array}{r@{\;}r@{}c@{}l}& 4x + 2y &=& 6 \\+& x - 2y &=& -1\end{array}\]
\(5x = 5\)
\(x = 1\)
Substitute x = 1 into Equation 1.\(2(1) + y = 3\)
\(2 + y = 3\)
\(y = 1\)
Answer: x = 1, y = 1
One equation equals 1. Rewrite 1 as a power of the same prime, then solve. To eliminate, multiply one equation by −1, then add.
Equation 1.\(8^x \cdot 2^y = 1\)
\((2^3)^x \cdot 2^y = 2^0\)
\(2^{3x} \cdot 2^y = 2^0\)
\(2^{3x+y} = 2^0\)
\(3x + y = 0\)
Equation 2.\(\dfrac{4^x}{2^y} = 32\)
\(\dfrac{(2^2)^x}{2^y} = 2^5\)
\(\dfrac{2^{2x}}{2^y} = 2^5\)
\(2^{2x-y} = 2^5\)
\(2x - y = 5\)
Add.\[\begin{array}{r@{\;}r@{}c@{}l}& 3x + y &=& 0 \\+& 2x - y &=& 5\end{array}\]
\(5x = 5\)
\(x = 1\)
Substitute x = 1 into Equation 1.\(3(1) + y = 0\)
\(3 + y = 0\)
\(y = -3\)
Answer: x = 1, y = −3
Equation 1.\(4^x \cdot 2^y = 1\)
\((2^2)^x \cdot 2^y = 2^0\)
\(2^{2x} \cdot 2^y = 2^0\)
\(2^{2x+y} = 2^0\)
\(2x + y = 0\)
Equation 2.\(\dfrac{8^x}{2^y} = 32\)
\(\dfrac{(2^3)^x}{2^y} = 2^5\)
\(\dfrac{2^{3x}}{2^y} = 2^5\)
\(2^{3x-y} = 2^5\)
\(3x - y = 5\)
Add.\[\begin{array}{r@{\;}r@{}c@{}l}& 2x + y &=& 0 \\+& 3x - y &=& 5\end{array}\]
\(5x = 5\)
\(x = 1\)
Substitute x = 1 into Equation 1.\(2(1) + y = 0\)
\(2 + y = 0\)
\(y = -2\)
Answer: x = 1, y = −2
Equation 1.\(9^x \cdot 3^y = 1\)
\((3^2)^x \cdot 3^y = 3^0\)
\(3^{2x} \cdot 3^y = 3^0\)
\(3^{2x+y} = 3^0\)
\(2x + y = 0\)
Equation 2.\(\dfrac{9^x}{3^y} = 81\)
\(\dfrac{(3^2)^x}{3^y} = 3^4\)
\(\dfrac{3^{2x}}{3^y} = 3^4\)
\(3^{2x-y} = 3^4\)
\(2x - y = 4\)
Add.\[\begin{array}{r@{\;}r@{}c@{}l}& 2x + y &=& 0 \\+& 2x - y &=& 4\end{array}\]
\(4x = 4\)
\(x = 1\)
Substitute x = 1 into Equation 1.\(2(1) + y = 0\)
\(2 + y = 0\)
\(y = -2\)
Answer: x = 1, y = −2
Equation 1.\(8^x \cdot 4^y = 1\)
\((2^3)^x \cdot (2^2)^y = 2^0\)
\(2^{3x} \cdot 2^{2y} = 2^0\)
\(2^{3x+2y} = 2^0\)
\(3x + 2y = 0\)
Equation 2.\(\dfrac{4^x}{2^y} = 128\)
\(\dfrac{(2^2)^x}{2^y} = 2^7\)
\(\dfrac{2^{2x}}{2^y} = 2^7\)
\(2^{2x-y} = 2^7\)
\(2x - y = 7\)
Multiply Equation 2 by 2.\(2(2x - y) = 14\)
\(4x - 2y = 14\)
Add.\[\begin{array}{r@{\;}r@{}c@{}l}& 3x + 2y &=& 0 \\+& 4x - 2y &=& 14\end{array}\]
\(7x = 14\)
\(x = 2\)
Substitute x = 2 into Equation 1.\(3(2) + 2y = 0\)
\(6 + 2y = 0\)
\(2y = -6\)
\(y = -3\)
Answer: x = 2, y = −3
Equation 1.\(3^x \cdot 9^y = 1\)
\(3^x \cdot (3^2)^y = 3^0\)
\(3^x \cdot 3^{2y} = 3^0\)
\(3^{x+2y} = 3^0\)
\(x + 2y = 0\)
Equation 2.\(\dfrac{9^x}{3^y} = 243\)
\(\dfrac{(3^2)^x}{3^y} = 3^5\)
\(\dfrac{3^{2x}}{3^y} = 3^5\)
\(3^{2x-y} = 3^5\)
\(2x - y = 5\)
Multiply Equation 1 by 2.\(2(x + 2y) = 0\)
\(2x + 4y = 0\)
Multiply 2x + 4y = 0 by −1.\(-(2x + 4y) = 0\)
\(-2x - 4y = 0\)
Add.\[\begin{array}{r@{\;}r@{}c@{}l}& 2x - y &=& 5 \\+& -2x - 4y &=& 0\end{array}\]
\(-5y = 5\)
\(y = -1\)
Substitute y = −1 into Equation 1.\(x + 2(-1) = 0\)
\(x - 2 = 0\)
\(x = 2\)
Answer: x = 2, y = −1
Equation 1.\(\dfrac{4^x}{8^y} = 1\)
\(\dfrac{(2^2)^x}{(2^3)^y} = 2^0\)
\(\dfrac{2^{2x}}{2^{3y}} = 2^0\)
\(2^{2x-3y} = 2^0\)
\(2x - 3y = 0\)
Equation 2.\(2^x \cdot 4^y = 128\)
\(2^x \cdot (2^2)^y = 2^7\)
\(2^x \cdot 2^{2y} = 2^7\)
\(2^{x+2y} = 2^7\)
\(x + 2y = 7\)
Multiply Equation 2 by 2.\(2(x + 2y) = 14\)
\(2x + 4y = 14\)
Multiply Equation 1 by −1.\(-(2x - 3y) = 0\)
\(-2x + 3y = 0\)
Add.\[\begin{array}{r@{\;}r@{}c@{}l}& 2x + 4y &=& 14 \\+& -2x + 3y &=& 0\end{array}\]
\(7y = 14\)
\(y = 2\)
Substitute y = 2 into Equation 2.\(x + 2(2) = 7\)
\(x + 4 = 7\)
\(x = 3\)
Answer: x = 3, y = 2
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