INDICIAL EQUATIONS

An indicial equation is the equations that involve the use of indices.

Example 1:  Solve for x in the equation 2^x = 32

Solution        2^x = 32

If you are given this type of question, it simply means you should express 32 in base 2. That is 2 raised to the power of what will give you 32. Therefore,

                        2^x = 2⁵

Since the base are the same, then

                        X = 5.

 Example 2:  Solve 3^(x+1) = 27

Solution        3^(x+1) = 27

Similarly express 27 as a base of 3. Therefore,

                        3^(x+1) = 3³

Since the base are the same, then

                        x + 1 = 3

                        x = 3 – 1 = 2.

Example 3: Solve 5^(x-3) =    1  

                                       25

Solution        5^(x-3) =  1  

                                 25

                                  

Note:  1   can be expressed as 25^-1. Then,

        25

         

            5^(x-3) = 25^-1

            5^(x-3) = (5²)^-1

            5^(x-3) = 5^-2

Since the base are the same, It can be cancel then we have

            x – 3 = -2

            x = -2 + 3

            x = 1.

Example 4: You should try this yourself, solve 2^(2x-3) = 16

If you are done that is very good, but did you get the answer? If otherwise

Check solution below

Solution        2^(2x-3) = 16

                        2^(2x-3) = 2⁴

                        2x – 3 = 4

                        2x = 4 + 3

                        2x = 7

                        x = 7/ 2

Example 5: Solve each of the following equations

• ·        (3^-1)^x = 27 
• ·        (0.125)^x+1 = 16        
• ·        5(5^x) = 125 
• ·        10^x = 0.000001 
• ·        5x = 40x^-1/2

Solution

·        (3^-1)^x = 27

                        3^-x = 3³

                        -x = 3

Multiply both sides by (-1). Then,

                        x = -3.

·        (0.125)^x+1 = 16

0.125 can be expressed as 125/1000Therefore, 125/1000 =1/8  =   (8^-1)

                                                                    

Then,                         (8)^-1(x+1) = 16

                        (2³)^(-x-1) = 2⁴

                        2^-3x-3 = 2⁴

                        -3x – 3 = 4

                        -3x = 4 + 3

                        -3x = 7 (by dividing through by -3)

                         x = 7/-3 

                             

·        5(5)^x = 125

Firstly, we divide through by 5, then

                        5^x = 25

                        5^x = 5²

                        x = 2

·        10^x = 0.000001

Firstly, we are to express 0.000001 as a fraction which is   1/1000000 = 10^-6

                        10^x = 10^-6

                        x = -6

·        5x = 40x^-1/2

Firstly, we are to divide through by 5. Because of the coefficient of x at the (LHS)  we need to make it a unity i.e  one (1) Then,

                        x = 8x^-1/2

Next is to divide through by x^-1/2. Then,

                  

              x/  x^-1/2     = 8

           

           

Recall, division law which states that if  
a^m÷aⁿ = a^m-n. Then,

                        X^{1- (-1/2)} = 8

                        X^1+1/2 = 8

                        X^3/2 = 8

Lastly, multiply the index through by the inverse of the power. i.e. 2/3

                        X^{3/2 * 2/3} = 8^2/3

                        X = 8^2/3

Recall, 8^2/3 is a form of power law. Then, it can be expressed as

X = (³√8)²

Cube root of 8 means a number, we can multiply by itself thrice or three times that will give 8. Answer is 2.

                        x = (2)²

                        x = 4.      

“Mathematicians are the priests of the modern world.”
― Bill Gaede.

So you can be a mathematician today. Just believe in yourself and you will be part of the priest of the modern world. Trust me it is just too easy… QEDMATHS