In an exponential equation the unknown is in the index, as in 2 x = 32 2^x = 32 2 x = 32 . The main method is one idea:
If a p = a q a^p = a^q a p = a q (with the same base a a a , not 0 or 1), then p = q p = q p = q .
So write both sides as powers of the same base , then set the indices equal. 2 x = 32 = 2 5 2^x = 32 = 2^5 2 x = 32 = 2 5 , so x = 5 x = 5 x = 5 .
Note
Useful powers to know: 2 5 = 32 2^5 = 32 2 5 = 32 , 2 6 = 64 2^6 = 64 2 6 = 64 , 2 7 = 128 2^7 = 128 2 7 = 128 , 3 4 = 81 3^4 = 81 3 4 = 81 , 3 5 = 243 3^5 = 243 3 5 = 243 , 5 3 = 125 5^3 = 125 5 3 = 125 , and 4 = 2 2 4 = 2^2 4 = 2 2 , 8 = 2 3 8 = 2^3 8 = 2 3 , 9 = 3 2 9 = 3^2 9 = 3 2 , 27 = 3 3 27 = 3^3 27 = 3 3 , 1 8 = 2 − 3 \frac18 = 2^{-3} 8 1 = 2 − 3 , 0.25 = 2 − 2 0.25 = 2^{-2} 0.25 = 2 − 2 .
More: make the bases the same
WAEC 2022 · Paper 1 · Q6 If 4 3 x = 16 x + 1 4^{3x} = 16^{x + 1} , find the value of x x . WAEC 2019 · Paper 1 · Q17 If ( 0.25 ) y = 32 (0.25)^y = 32 , find the value of y y . JAMB 2003 · UME · Q11 If 9 2 x − 1 27 x + 1 = 1 \dfrac{9^{2x - 1}}{27^{x + 1}} = 1 , find the value of x x . JAMB 1997 · UME · Q6 If 8 x 2 = 2 3 8 × 4 3 4 8^{\frac x2} = 2^{\frac38} \times 4^{\frac34} , find x x . JAMB 1993 · UME · Q3 If 9 x − 1 2 = 3 x 2 9^{x - \frac12} = 3^{x^2} , find the value of x x . JAMB 1983 · UME · Q32 If ( 2 3 ) m ( 3 4 ) n = 256 729 \left(\frac23\right)^m\left(\frac34\right)^n = \frac{256}{729} , find the values of m m and n n . JAMB 1989 · UME · Q17 Find the positive number x x such that 2 x 3 − x 2 − 2 x = 1 2^{x^3 - x^2 - 2x} = 1 . WAEC 2019 · Paper 2 · Q3 Ali, Musah and Yusif shared ₦420,000.00 in the ratio 3 : 5 : 8 3 : 5 : 8 respectively. Find the sum of Ali's and Yusif's shares. WAEC 2019 · Paper 2 · Q2 The slant height of a cone is 18.7 cm 18.7\text{ cm} and the diameter is 24 cm 24\text{ cm} . Calculate, correct to three significant … JAMB 1992 · UME · Q4 What is the value of x x satisfying the equation 4 2 x 4 3 x = 2 \frac{4^{2x}}{4^{3x}} = 2 ? JAMB 1993 · UME · Q4 Solve for y y in the equation 10 y × 5 2 y − 2 × 4 y − 1 = 1 10^y \times 5^{2y - 2} \times 4^{y - 1} = 1 . WAEC 2009 · Paper 2 · Q1 Given that ( 3 − 5 2 ) ( 3 + 2 ) = a + b 6 (\sqrt3 - 5\sqrt2)(\sqrt3 + \sqrt2) = a + b\sqrt6 , find a a and b b . NECO 2022 · Paper 1 · Q4 If 3 ( 6 − 9 x ) = 27 ( 1 − 2 x ) 3^{(6 - 9x)} = 27^{(1 - 2x)} , find the value of x x .
Worked example · NECO 2023
NECO 2023 · Paper 2 · Q4 (a)
Solve the equation 27 2 x − 1 × ( 1 3 ) − ( 3 x + 2 ) = 9 x + 3 27^{2x - 1} \times \left(\frac13\right)^{-(3x + 2)} = 9^{x + 3} 2 7 2 x − 1 × ( 3 1 ) − ( 3 x + 2 ) = 9 x + 3 .
Everything as a power of 3
27 = 3 3 27 = 3^3 27 = 3 3 , 1 3 = 3 − 1 \frac13 = 3^{-1} 3 1 = 3 − 1 and 9 = 3 2 9 = 3^2 9 = 3 2 .
Think first. Write 27, 1 3 \frac13 3 1 and 9 as powers of 3.
Multiply out the indices
27 2 x − 1 = 3 6 x − 3 27^{2x - 1} = 3^{6x - 3} 2 7 2 x − 1 = 3 6 x − 3 , ( 1 3 ) − ( 3 x + 2 ) = 3 3 x + 2 \left(\frac13\right)^{-(3x + 2)} = 3^{3x + 2} ( 3 1 ) − ( 3 x + 2 ) = 3 3 x + 2 and 9 x + 3 = 3 2 x + 6 9^{x + 3} = 3^{2x + 6} 9 x + 3 = 3 2 x + 6 .
Add the indices on the left
3 ( 6 x − 3 ) + ( 3 x + 2 ) = 3 9 x − 1 3^{(6x - 3) + (3x + 2)} = 3^{9x - 1} 3 ( 6 x − 3 ) + ( 3 x + 2 ) = 3 9 x − 1 , so 3 9 x − 1 = 3 2 x + 6 3^{9x - 1} = 3^{2x + 6} 3 9 x − 1 = 3 2 x + 6 .
Think first. Multiplying powers of 3: what happens to the indices?
Compare indices
9 x − 1 = 2 x + 6 9x - 1 = 2x + 6 9 x − 1 = 2 x + 6 , so 7 x = 7 7x = 7 7 x = 7 and x = 1 x = 1 x = 1 .
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When the terms are added: factor out the power
3 x + 3 x + 1 3^x + 3^{x + 1} 3 x + 3 x + 1 can’t be combined into one power directly, but 3 x + 1 = 3 × 3 x 3^{x + 1} = 3 \times 3^x 3 x + 1 = 3 × 3 x . Take 3 x 3^x 3 x out as a common factor.
More: factor out the power
Equations that hide a quadratic
4 x = ( 2 2 ) x = ( 2 x ) 2 4^x = (2^2)^x = (2^x)^2 4 x = ( 2 2 ) x = ( 2 x ) 2 , so an equation with 4 x 4^x 4 x and 2 x 2^x 2 x in it is a quadratic↺ in disguise. Write y = 2 x y = 2^x y = 2 x , solve for y y y , then turn each y y y back into x x x . For 4 x − 6 ( 2 x ) + 8 = 0 4^x - 6(2^x) + 8 = 0 4 x − 6 ( 2 x ) + 8 = 0 :
4ˣ = (2²)ˣ = (2ˣ )² (2ˣ )² − 6(2ˣ ) + 8 = 0 y ² − 6y + 8 = 0A quadratic in disguise Let y = 2ˣ
y 2 − 6 y + 8 = 0 y^2 - 6y + 8 = 0 y 2 − 6 y + 8 = 0 gives ( y − 2 ) ( y − 4 ) = 0 (y - 2)(y - 4) = 0 ( y − 2 ) ( y − 4 ) = 0 , so 2 x = 2 2^x = 2 2 x = 2 or 2 x = 4 2^x = 4 2 x = 4 , and x = 1 x = 1 x = 1 or x = 2 x = 2 x = 2 . A negative value of y y y gives no answer, because 2 x 2^x 2 x is always positive.
When the unknown is the base
In x 2 3 = 9 x^{\frac23} = 9 x 3 2 = 9 the unknown is the base , not the index. Undo the power by raising both sides to its reciprocal: x = 9 3 2 = 27 x = 9^{\frac32} = 27 x = 9 2 3 = 27 . A root works the same way: from 2 + x + 1 3 = 5 2 + \sqrt[3]{x + 1} = 5 2 + 3 x + 1 = 5 , get the root alone (x + 1 3 = 3 \sqrt[3]{x + 1} = 3 3 x + 1 = 3 ), then cube both sides (x + 1 = 27 x + 1 = 27 x + 1 = 27 , so x = 26 x = 26 x = 26 ).
More: the unknown in the base
Two equations together
When two exponential equations share two unknowns, turn each into a simple equation in the indices, then solve the pair as simultaneous equations↺ .
Worked example · WAEC 2019
WAEC 2019 · Paper 2 · Q6 (a)
Given that 2 m × ( 1 8 ) n = 128 2^m \times \left(\frac18\right)^n = 128 2 m × ( 8 1 ) n = 128 and 4 m ÷ 2 − 4 n = 1 16 4^m \div 2^{-4n} = \frac{1}{16} 4 m ÷ 2 − 4 n = 16 1 , find the value of ( m − n ) (m - n) ( m − n ) .
The first equation in base 2
2 m × 2 − 3 n = 2 7 2^m \times 2^{-3n} = 2^7 2 m × 2 − 3 n = 2 7 , so m − 3 n = 7 m - 3n = 7 m − 3 n = 7 .
Think first. 1 8 = 2 ? \frac18 = 2^{?} 8 1 = 2 ? and 128 = 2 ? 128 = 2^{?} 128 = 2 ?
The second equation in base 2
4 m = 2 2 m 4^m = 2^{2m} 4 m = 2 2 m , and dividing by 2 − 4 n 2^{-4n} 2 − 4 n adds 4 n 4n 4 n : 2 2 m + 4 n = 2 − 4 2^{2m + 4n} = 2^{-4} 2 2 m + 4 n = 2 − 4 , so 2 m + 4 n = − 4 2m + 4n = -4 2 m + 4 n = − 4 , or m + 2 n = − 2 m + 2n = -2 m + 2 n = − 2 .
Solve the pair
Subtract: ( m − 3 n ) − ( m + 2 n ) = 7 − ( − 2 ) (m - 3n) - (m + 2n) = 7 - (-2) ( m − 3 n ) − ( m + 2 n ) = 7 − ( − 2 ) , so − 5 n = 9 -5n = 9 − 5 n = 9 and n = − 9 5 n = -\frac95 n = − 5 9 . Then m = 7 + 3 n = 8 5 m = 7 + 3n = \frac85 m = 7 + 3 n = 5 8 .
Answer the question asked
m − n = 8 5 + 9 5 = 17 5 = 3 2 5 m - n = \frac85 + \frac95 = \frac{17}{5} = 3\frac25 m − n = 5 8 + 5 9 = 5 17 = 3 5 2 .
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More: two equations together
Your turn
If 16 × 2 ( x + 1 ) = 4 x × 8 ( 1 − x ) 16 \times 2^{(x + 1)} = 4^x \times 8^{(1 - x)} 16 × 2 ( x + 1 ) = 4 x × 8 ( 1 − x ) , find the value of x x x .
Worked solution (try it first) Write everything as a power of 2.
Left:
16 × 2 x + 1 = 2 4 × 2 x + 1 16 \times 2^{x + 1} = 2^4 \times 2^{x + 1} 16 × 2 x + 1 = 2 4 × 2 x + 1 Right:
4 x = 2 2 x 4^x = 2^{2x} 4 x = 2 2 x and
8 1 − x = 2 3 − 3 x 8^{1 - x} = 2^{3 - 3x} 8 1 − x = 2 3 − 3 x , so the right side is
2 2 x + 3 − 3 x = 2 3 − x 2^{2x + 3 - 3x} = 2^{3 - x} 2 2 x + 3 − 3 x = 2 3 − x .
Equate the powers:
x + 5 = 3 − x x + 5 = 3 - x x + 5 = 3 − x , so
2 x = − 2 2x = -2 2 x = − 2 and
x = − 1 x = -1 x = − 1 , option D.
Watch out
Multiply the whole bracket: 8 1 − x = 2 3 ( 1 − x ) = 2 3 − 3 x 8^{1 - x} = 2^{3(1 - x)} = 2^{3 - 3x} 8 1 − x = 2 3 ( 1 − x ) = 2 3 − 3 x . Writing 2 3 − x 2^{3 - x} 2 3 − x gives x + 5 = 3 + x x + 5 = 3 + x x + 5 = 3 + x , which has no solution. Report a problem with this question
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