Exponential Equation
An equation which contains the unknown variable appearing as an exponent of a base is known as an exponential equation. In the equation 5x = 25, the unknown variable x is an exponent of base 5. So this equation is called an exponential equation.
The axioms given below help to solve the
exponential equations:
a)
If xa = ya,
then x = y
b)
If xa=xb,
then a = b
c)
If xa = 1, then xa
= x0. So a = 0
The following steps are useful to solve
the exponential equation:
1.
Simplify both sides of the
equation
2.
Make both sides of the
equation into the same base.
3.
Equate their exponents and
simplify.
Workout Examples
Solution: Here,
a)
2x+4 = 8x
or, 2x+4
= (23)x
or, 2x+4
= 23x
∴ x+4 = 3x
or, 4 = 3x – x
or, 4 = 2x
or, x = 4/2
or, x = 2
Example 3: Solve: 32x+1 – 9x+1 + 54 = 0
Solution: Here,
32x+1 – 9x+1
+ 54 = 0
or, 32x.31
– 9x.91 = -54
or, 9x.3
– 9x.9 = -54
or, 9x(3
– 9) = -54
or, 9x.-6
= -54
or, 9x
= -54/-6
or, 9x
= 9
or, 9x
= 91
∴ x = 1
or, 25x2
+ 25 = 626x
or, 25x2
– 626x + 25 = 0
or, 25x2
– 625x – x + 25 = 0
or, 25x(x – 25) –
1(x – 25) = 0
or, (x – 25)(25x
– 1) = 0
∴ Either, x – 25 = 0
or,
x = 25
or,
5a = 52
∴ a = 2
Or, 25x – 1 = 0
or,
25x = 1
or,
x = 1/25
or,
x = 1/52
or,
5a = 5-2
∴ a = -2
∴ a = 2 or -2
Example 7: Solve: 4 × 3x+1 =
27 + 9x
Solution: Here,
4 × 3x+1 = 27 + 9x
or, 9x
– 4 × 3x+1 + 27 = 0
or, 32x
– 4 . 3x . 31 + 27 = 0
or, (3x)2
– 12 . 3x + 27 = 0
Let, 3x = a
Then, a2
– 12a + 27 = 0
or, a2
– 9a – 3a + 27 = 0
or, a(a
– 9) -3(a – 9) = 0
or, (a
– 9)(a – 3) = 0
∴ Either, a – 9 = 0
or,
a = 9
or,
3x = 32
∴ x = 2
Or, a – 3 = 0
or,
a = 3
or,
3x = 31
∴ x = 1
∴ x = 1 or 2
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questions or problems regarding the exponential equations here.
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