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Exponential equations of the form k a ( x + p ) = m

Exponential equations generally have the unknown variable as the power. The following are examples of exponential equations:

2 x = 1 2 - x 3 x + 1 = 2 4 3 - 6 = 7 x + 2

You should already be familiar with exponential notation. Solving exponential equations is simple, if we remember how to apply the laws of exponentials.

Investigation : solving exponential equations

Solve the following equations by completing the table:

2 x = 2 x
-3 -2 -1 0 1 2 3
2 x
3 x = 9 x
-3 -2 -1 0 1 2 3
3 x
2 x + 1 = 8 x
-3 -2 -1 0 1 2 3
2 x + 1

Algebraic solution

Equality for Exponential Functions

If a is a positive number such that a > 0 , (except when a = 1 ) then:

a x = a y

if and only if:

x = y
(If a = 1 , then x and y can differ)

This means that if we can write all terms in an equation with the same base, we can solve the exponential equations by equating the indices. For example takethe equation 3 x + 1 = 9 . This can be written as:

3 x + 1 = 3 2 .

Since the bases are equal (to 3), we know that the exponents must also be equal. Therefore we can write:

x + 1 = 2 .

This gives:

x = 1 .

Method: solving exponential equations

  1. Try to write all terms with the same base.
  2. Equate the exponents of the bases of the left and right hand side of the equation.
  3. Solve the equation obtained in the previous step.
  4. Check the solutions

Investigation : exponential numbers

Write the following with the same base. The base is the first in the list. For example, in the list 2, 4, 8, thebase is two and we can write 4 as 2 2 .

  1. 2,4,8,16,32,64,128,512,1024
  2. 3,9,27,81,243
  3. 5,25,125,625
  4. 13,169
  5. 2 x , 4 x 2 , 8 x 3 , 49 x 8

Solve for x : 2 x = 2

  1. All terms are written with the same base.

    2 x = 2 1
  2. x = 1
  3. LHS = 2 x = 2 1 = 2 RHS = 2 1 = 2 = LHS

    Since both sides are equal, the answer is correct.

  4. x = 1

    is the solution to 2 x = 2 .

Solve:

2 x + 4 = 4 2 x
  1. 2 x + 4 = 4 2 x 2 x + 4 = 2 2 ( 2 x ) 2 x + 4 = 2 4 x
  2. x + 4 = 4 x
  3. x + 4 = 4 x x - 4 x = - 4 - 3 x = - 4 x = - 4 - 3 x = 4 3
  4. LHS = 2 x + 4 = 2 ( 4 3 + 4 ) = 2 16 3 = ( 2 16 ) 1 3 RHS = 4 2 x = 4 2 ( 4 3 ) = 4 8 3 = ( 4 8 ) 1 3 = ( ( 2 2 ) 8 ) 1 3 = ( 2 16 ) 1 3 = LHS

    Since both sides are equal, the answer is correct.

  5. x = 4 3

    is the solution to 2 x + 4 = 4 2 x .

Solving exponential equations

  1. Solve the following exponential equations.
    a. 2 x + 5 = 2 5 b. 3 2 x + 1 = 3 3 c. 5 2 x + 2 = 5 3
    d. 6 5 - x = 6 12 e. 64 x + 1 = 16 2 x + 5 f. 125 x = 5
  2. Solve: 3 9 x - 2 = 27
  3. Solve for k : 81 k + 2 = 27 k + 4
  4. The growth of an algae in a pond can be modeled by the function f ( t ) = 2 t . Find the value of t such that f ( t ) = 128
  5. Solve for x : 25 ( 1 - 2 x ) = 5 4
  6. Solve for x : 27 x × 9 x - 2 = 1

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Source:  OpenStax, Siyavula textbooks: grade 10 maths [ncs]. OpenStax CNX. Aug 05, 2011 Download for free at http://cnx.org/content/col11239/1.2
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