Indices

Base vs Power

Remember the following rules:

Same base:

a^m x aⁿ = a ^(m+n)

a^m / aⁿ = a ^(m-n)

Same power

a^m x b^m = (ab) ^m

a^m / b^m = (a/b) ^m

Same base and same power then you can add or subtract:

2a^m + a^m =  a^m (2+1) =3a^m

2a^m - a^m =  a^m (2-1) =a^m

^{Solving indices equations type 1 --> There are only 2 non- zero terms, and they can be written into two with the same base.}

E.g. Solve 2^x = 8 ^x-3

Step 1: Are there 2 non- zero terms only?  Yes

Step 2: Can they be written into terms with the same base? Yes, as 8 can be written as 2³

Step 3: Write them into the same base, and compare the power.

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

x = 3(x-3)

x = 3x -3

x – 3x = -3

-2x = -3

x = 1.5

Solving indices equations type 2 --> There are only 2 non- zero terms, and they CANNOT be written into two with the same base.

2^x = 5 ^2x-3

Step 1: Are there 2 non- zero terms only?  Yes

Step 2: Can they be written into terms with the same base? No. 5 and 2 cannot be simplified to a number with the same base.

Step 3: ln both sides, and make x the power.

2^x = 5 ^2x-3

ln2^x = ln5 ^2x-3

xln2 = (2x-3)ln5

xln2 =2xln5 – 3ln5

2xln5+xln2 = 3ln5

x(2ln5 +ln2) = 3ln5

x = 3ln5/(2ln5+ln2) =1.23

Solving indices equations type 3 --> You can simplify to 3 non- zero terms. 2 of the terms have the same base, and one has twice the power of the other.

4^x -2.2^x +1 =0    ----- (1)

4^x  = 2^2x

Let y = 2^x, and so y² = 2^2x.

Equation (1) become:

y² -2y+1 =0

(y-1)² = 0

y=1, remember to substitute back y = 2^x, since you are solving for x.

2^x =1 à this is solving using type 1 method, as 1 = 2⁰.

2^x = 2⁰

x=0