First, consider this practice problem:

- \(7^6 + 7^8\)

A. \(7^{14}\)

B. \(7^{16}\)

C. \(7^{48}\)

D. \(14^{14}\)

E. \((50)times(7^6)\)

If this problem makes your head spin, you have found the right blog article!

## Operations with powers

There are legitimate laws of exponents for multiplying and dividing powers.

Take a look at these examples:

- \(atimes(b + c) = atimes b + atimes c\)
- \((a^m) times (a^n) = a^{m + n}\)
- \( frac{a^m}{a^n} = a^{m – n}\)
- \((7^8) times (7^6) = 7^{8+6} = 7^{14}\)
- \( frac{7^8}{7^6} = 7^{8 – 6} = 7^2 = 49\)

Multiplying powers means add the exponents and dividing powers means subtract the exponents. For more on this, see this post on exponent properties.

Alas! There is no law of exponents for adding and subtracting powers. 😢

In other words, \(a^n\) plus or minus \(a^m\) is not going to equal a-to-the-power-of-anything-in-particular.

There is no convenient way to combine a sum or difference of powers into a single power expression.

## Distributing and factoring out

But, all is not lost! We can avail ourselves of one of the fundamental number properties, the Distributive Law, and its inverse process, “factoring out.”

That is a rather remarkable and often underestimated equation.

First of all, it is a statement of the Distributive Law. Remember \(atimes(b + c) = atimes b + atimes c\)?

If we start on the left and go to the right, we are

distributing“a.” The reverse process is calledfactoring out: if we start on the right side, and go to the left, we are factoring “a” out.

Multiplication distributes over addition and subtraction: we could replace both plus signs with minus signs in that equation and it would still be true.

Written in simple variable form, it probably doesn’t look all that earth-shattering, but the implications of that equation are so widespread as to boggle the mind. Addition and subtraction of variables, addition and subtraction of fractions, and all techniques for factoring quadratics are ultimately dependent on this profound fundamental law.

Before discussing the implications for addition and subtraction of powers, I will just make one more point. In this above equation, “a” can be any common factor of the two terms on the right, but in some sense, we “maximize” the effect of factoring out when the factor we factor out is the Greatest Common Factor (the GCF). See this link for a review of how GCF works with numbers.

## Factoring out to add & subtract powers

Suppose we want to simplify \(2^{17} – 2^{13} = ?\)

Well, since both terms are powers of two, they share several common factors (\(2^1\), \(2^2\), \(2^3\), etc.).

The GCF of both of these terms is \(2^{13}\) itself. We can express each term as the product of the GCF and something else.

\(2^{17} = (2^4) times (2^{13})\)\(2^{13} = (1) times (2^{13})\)

Now, we can write the difference of powers as

\(2^{17} – 2^{13} = (2^4)times (2^{13}) – (1)times (2^{13}) = ?\)I realize that, at the moment, it may not be completely obvious that this will ultimately be a step in the direction of *simplification* — it looks like we made everything gratuitously more complex in the second step! Trust that things will simplify soon! Here’s the Distributive Law equation again, now with subtraction and with left & right switched from above.

Notice that the second step of the power subtraction equation matches perfectly the pattern on the left of this statement of the Distributive Law: \(a = 2^{13}\), \(b = 2^4\) and \(c = 1\). Following this pattern, we see this turns into:

\(2^{17} – 2^{13} = (2^4)times (2^{13}) – (1)times (2^{13}) = (2^{13})times [(2^4) – 1]\)Now, it’s germane to point out: we should be able to simplify \(2^4\).

\(2^1 = 1\)

\(2^2 = 4\)

\(2^3 = 8\)

\(2^4 = 16\)

That, in turn, allows us to simplify the piece in the brackets in the last expression:

\(2^{17} – 2^{13} = (2^4)times (2^{13}) – (1)times (2^{13}) = (2^{13})times [(2^4) – 1]\)

\(= (2^{13})times [16 – 1] = (2^{13})times (15) \)

Voila! That final expression, fifteen times two to the thirteen, is the simplification of the difference of two powers.

This process would be particularly important if, for example, you were asked to simplify something like:

\(frac{[2^{17} – 2^{13}]}{20}\)We can’t cancel from either term in the numerator as long as we have addition or subtraction in the numerator, but once we simplify the numerator so that it’s all multiplication, we can cancel. You should verify for yourself that:

\(frac{[2^{17} – 2^{13}]}{20} = (3)times (2^{11})\)A remarkable simplification!

## Further Practice

Now that you have seen that, take another look at the practice question from above, before reading the answer and explanation below.

Practice question #1

Remember the first practice question you encountered before this article began?

- \(7^6 + 7^8\)

A. \(7^{14}\)

B. \(7^{16}\)

C. \(7^{48}\)

D. \(14^{14}\)

E. \((50)times(7^6)\)

## Click here for the answer and explanation to practice question #1

The GCF of \(7^6 \)and \(7^8\) is \(7^6\).

\(7^6 = (1)times (7^6)\)

\(7^8 = (7^2)times (7^6)\)

Therefore

\(7^6 + 7^8 = (1)times (7^6) + (7^2)times (7^6) = [1 + 7^2]times (7^6) \)

\(= [1 + 49]times (7^6)\)

\(= (50)times (7^6)\)

Answer = **E**

Did your answer change after you read this post? Let us know in the comments below.

Practice question #2

- if \(2^{2n} + 2^{2n} + 2^{2n} + 2^{2n} = 4^{24}\), then n

A. 3

B. 6

C. 12

D. 23

E. 24

▶ Click here for the answer and explanation to practice question #2

4power m – 4power m-1=24.find m. My answer came 2.5

Assuming you mean (4^m) – [4^(m-1)], then yes, 2.5 would be correct. 🙂 This process, of course, involves non-integer powers, a complicated operation that isn’t really covered in this post. Quora has a good thread on these kinds of exponents here.

pls solve it

you can write A to the power B as A^B and A divided by B as A/B, and A multiplied by B as A*B, etc

Can you solve this

2×10^6-1.2×10^-3

Yes, I can solve it. 2*10^6 would be 2,000,000, and 1.2*10^-3 is 1.2*(10/10/10), or 1.2*0.001, or. From there, it’s a matter of simple subtraction. Assuming of course, that you meant the order of operations to look as follows: (2×10^6)-(1.2×10^-3). Otherwise, without the parenthesis, your default order of operations would lead to a more complicated equation and a different answer. 😉

Can you solve (T^-1 + T^2 +T^1)

Hi Hitesh,

Can you provide some more detail about where you found this question and the context of the question? We are not able to provide support for outside materials, but we are more than happy to help with questions related to Magoosh and Official materials. For this question, it’s important to realize that you will be dealing with a polynomial expression that will have multiple answers. You will need to solve this using a quadratic equation. As we say in this blog post, there is no ‘easy’ method for solving questions that involve addition and subtraction of powers.

Pls solve it

(1/6)^2 +(1/6)^4 =6^2(m+1)

Than find m

An interesting equation, Kajolsharma. 🙂 A couple of questions before we look at a solution, though. First, where did you see this equation? Second, is (m+1) supposed to be part of the exponential notation, or is it supposed to be a separate expression, multiplied by (6^2). In other words, does the equation mean this:

(1/6)^2 +(1/6)^4 =6^[2(m+1)]

or this:

(1/6)^2 +(1/6)^4 =(6^2)(m+1)

With your current notation, things are a little bit ambiguous, so let me know. 😉

BTW what I did was to substitute 10 for 2 to start.

and the verbiage above SHOULD have been more than, less than or equal to.

sorry.

How do you do:

4^3+4^3+4^3+4^3

I couldn’t work it out, please answer so I can understand a bit better!

Cheers, Sudhan

Hi Sudhan,

We can find the sum of 4^3+4^3+4^3+4^3 in a couple of ways. For example, recognizing that we have the same term, 4^3, repeated 4 times, we can rewrite the expression as 4*(4^3). From there, we will want to evaluate 4^3 = 64 and find the product

4*64 = 256.

We could also first evaluate 4^3 = 64 and add replace each term of 4^3 with 64 to find the sum:

4^3+4^3+4^3+4^3 = 64+64+64+64 = 256.

I hope this helps 🙂

I am unable to solve below.

2^x+y * 3^x-y = 32/3

Im finding it difficult to get from (2^17-2^13/20) to the answer of (3)*(2^11). I might be doing something wrong in my process but cant figure out how you simplified to 3 and why the change from 2^13 to 2^11.

Neka

In the post above, I showed how 2^17-2^13 simplifies to (15)*(2^13). I’ll assume you follow that, because the steps are printed above. The denominator factors into

20 = 5*4 = 5*(2^2), When we divide (15)*(2^13) by 5*(2^2), the 15/5 =3, and the (2^13)/(2^2) = 2^11.

Does all this make sense?

Mike 🙂

Awesome. Thanks. Didn’t realize I could break down the denominator.

Neka,

You are quite welcome, my friend. Best of luck to you!

Mike 🙂

That was so helpful, and not in my Kaplan book at all. Thank you!

Dear Kate,

I’m very glad you found this helpful. Best of luck to you!

Mike 🙂

7^6 (1 + 7^2) —-> 7^6 ( 1+ 49) —-> 7^6 * 50

10 seconds 😉

Thanks Mike

You are quite welcome.

Mike 🙂