Exponential Equation

Exponential Equation

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.

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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

Example 1: Solve the following exponential equations.  a) 2x+4 = 8x   b) 9^((x+5)/2) =〖81〗^((x+1)/2)   c) 5x = 1/(0.04)

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 

      

b) 9^((x+5)/2) = 〖81〗^((x+1)/2)

           

c) 5x = 1/(0.04)

Example 2: Solve: ax-2.b = bx-2.a


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

 

Example 4: If xa = yb = zc and y = √xz , prove that 2/b = 1/a+1/c




Example 5: If x = 3^(1/3)+3^((-1)/3), prove that 3x3 – 9x – 10 = 0

 

Example 6: Solve: 5a + 1/5^a  = 251/25   Solution: Let, 5a = x

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

 

You can comment your questions or problems regarding the exponential equations here.


1 comment:

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