Exponential (Indicial) Equations

Exponential or Indicial Equation is a combination of indices and all other forms of equations, it is very easy to solve provided you have excellent knowledge of the laws of Indices.

Rules for Solving Exponential (Indicial) Equations

1. The two sides i.e LHS and RHS of the equation must be expressed in index form.

2. The two sides of the equation must also have the same values for you to cancel them out.

3. You’ll always solve for an unknown value which can be represented by any letter of the alphabet.

Note

You will need to master all the laws of indices, if you must properly understand exponential equations.

Examples

1. If 3² = 3², find x.

Solution

3² = 3²

Step 1

Note that the equations above are already expressed in index form, so just cancel the similar ones out;

3 cancels 3.

 x = 2.

2. If 2^{x + 1} = 2³, find x.

Solution

2^{x + 1} = 2³

Step 1

The equation is already in index form, so just cancel out the similar ones;

2 cancels 2;

x + 1 = 3

Step 2

Make x the subject by carrying +1 to the RHS;

x = 3 – 1

x = 2.

3. If 3^x = 9, solve for x.

Solution

3^x = 9

Step 1

Express 9 in index form; expressing 9 in index form is 3² which gives (3  3).

3² = 3²

Step 2

3 cancels 3;

x = 2.

4. 16^x = 0.125. Solve for x.

Solution

16^x = 0.125

Step 1

Express 0.125 in fraction by carrying the points out;

Step 2

Cancellation Process;

5 goes in 125 gives 25 while 5 goes in 1000 gives 200;

Cancellation Process;

5 goes in 25 gives 5 while 5 goes in 200 gives 40;

Cancellation Process;

5 goes in 5 gives 1 while 5 goes in 40 gives 8;

Step 3

Apply Negative index law to the fraction at the right;

16^x = 8^-1

Step 4

Take both values to index form;

2^4(x) = 2^3(-1)

Step 5

2 cancels 2;

4(x) = 3(-1)

Step 6

Remove the brackets;

4x = -3

Step 7

Divide both sides by 4;

Cancellation Process;

4 cancels 4;

5. 8(4^x) = 32, solve for x.

Solution

8(4^x) = 32

Step 1

Remove the bracket at the LHS;

8 x 4^x = 32

32^x = 32

Step 2

32 cancels 32;

x = 1.

6.

Add caption

Solution

Step 1

Apply negative index law to the fraction at the RHS;
2^x(x-3) = 4^-1

Step 2

Express the value 4^-1 in index from;

2^x(x-3) = 2^2(-1)

Step 3

2 cancels 2;

x(x – 3) = 2(-1)

Step 4

Remove the brackets;

x² - 3x = -2

Step 5

Rearrange;

x² - 3x + 2 = 0.  ----quadratic equation.

As you can see, this is a quadratic equation. So let’s factorise!

x² - 3x + 2 = 0         
                                     +2x²   (x)    

                                     -3x    (+)

                                     (-2x & -1x)

x² - 2x – 1x + 2 = 0

x(x – 2) – 1(x – 2) = 0

(x – 2)(x – 1) = 0

(x – 2) = 0 OR (x – 1) = 0

x – 2 = 0 OR x – 1 = 0

x = 0 + 2 OR x = 0 + 1

x = 2 OR x = 1.

7. 2^2(x-3) = 1. Solve for x.

Solution

2^2(x-3) = 1

Step 1

Apply zero index law to 1;

2^2(x-3) = 2⁰

Step 2

2 cancels 2;

2(x – 3) = 0

Step 3

Remove bracket;

2x – 6 = 0

Step 4

Rearrange;

2x = 0 + 6

2x = 6

Step 5

Divide both sides by 2;

Cancellation Process;

2 cancels 2 while 2 goes in 6 gives 3;

x = 3.

Assignment

Solve the following equations;

1. 27^x = 81

2. 32^x = 0.0625

3. 4^6(x-3) = 1

4. 16(4^x) = 64

6. 4^{x + 2} = 8^{x – 1}

7. 8^x = 0.25

9. 2^x = 16

10. 2^{3x - 1} = 4^{x + 3}