Adding terms with exponents

An SAT Math micro-topic under Radicals and rational exponents (Advanced Math). Free to read — no account needed.

Adding powers is a common trap.
You add exponents only when you multiply like bases (

2^3 \times 2^4 = 2^7

).
When you add them, the exponents do not combine —

2^{10} + 2^{10}

is not

2^{20}

When the two powers are equal, adding them just doubles one.
So

2^{10} + 2^{10} = 2 \times 2^{10}

, and since

2 = 2^1

, that is

2^{11}

.
In the same way, three copies would give

3 \times 2^{10}

When the powers differ, factor out the smaller one.
For

3^{50} - 3^{48}

, pull out

3^{48}

3^{50} - 3^{48} = 3^{48}(3^2 - 1)

.
Then

3^2 - 1 = 8

, so the whole thing is

8 \times 3^{48}

This shows up in evenly spaced sequences, where each term is a fixed power apart.
If

S_7 - S_6 = 2^{10}

and

S_6 = 2^{10}

, then

S_7 = 2^{10} + 2^{10} = 2^{11}

.
Each new term is found by adding, then factoring out the common power.

Worked examples

Simplify

5^8 + 5^8

.
The two powers are equal, so adding them doubles one:

5^8 + 5^8 = 2 \times 5^8

.
So the answer is

2 \times 5^8

Simplify

2^{20} - 2^{18}

.
Factor out the smaller power,

2^{18}

2^{20} - 2^{18} = 2^{18}(2^2 - 1)

.
Since

2^2 - 1 = 3

, the result is

3 \times 2^{18}

In a sequence,

S_8 - S_7 = 2^{10}

and

S_7 = 2^{11}

. What is

S_8

?
Add:

S_8 = 2^{11} + 2^{10}

, and factor out

2^{10}

to get

2^{10}(2 + 1)

.
So

S_8 = 3 \times 2^{10}

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