Can you show me how [math]2^k + 2^k = 2(2^k) [/math]?


These sort of proofs are done with PMI (Principle of Mathematical Induction).

According to PMI, We can proof this equation in three steps:

  1. Prove this equation for k=1. LHS=[math](2^1)+(2^1)=2+2=4[/math]. RHS=[math]2(2^1)=2*2=4[/math]
  2. Prove this equation for k=n (some real number n) . LHS=[math](2^n)+(2^n)[/math] RHS=[math]2(2^n)[/math] , This is already shown in question.
  3. Prove this equation for k=n+1 (some real number n) . LHS= [math](2^(n+1))+(2^(n+1))=2(2^n)+2(2^n)=2[2^n+2^n][/math]. Using result of step 2 shown above : =[math]2[2(2^n)][/math]. RHS= [math]2(2^(n+1))=2[2(2^n)][/math] .

LHS=RHS. Proved

Hence, by Principle of Mathematical Induction this equation holds true for all real values of k.

19 Replies to “Can you show me how [math]2^k + 2^k = 2(2^k) [/math]?”

  1. Let’s take a step back from your problem, so we can see it more clearly for what it really is. To understand how this equation is true, is to understand what some mathematical “language” means.

    Mathematical notation is, firstly, a way of writing out some rather long sentences with many fewer words. It came about because people got tired of saying things like “Think of a number, then multiply two by itself a number of times equal to the number you first thought of, then add the result to itself.” Instead of that sentence, we can just write “k” for “the number you first thought of”, raise 2 to the power k in place of “multiply two by itself a number of times equal to” k, and write “+” for “add the result to itself”. Even our Indo-Arabic number notation, which lets us write “2” for “two”, is just another short-cut, but a very convenient one.

    And maths notation is more than that – it’s also become an international language for the same ideas. No matter whether you speak English, German, Russian, Chinese or Greek, you can write down the same maths ideas in exactly the same way, yet read them in your own language. How cool is that?

    So our first task in solving your maths problem is to understand what this “sentence” is saying. We’ve actually already explained what the stuff before the equals sign “=”, the left-hand side (or LHS), means. So what does the RHS (right hand side) mean?

    Whenever we write an expression like a(b), we’re using parentheses – “(”, left parenthesis and “)”, right parenthesis – which we sometimes loosely call “brackets” – to group stuff together. In the expression a(b), the parentheses just make sure that we deal with whatever b is before we worry about a. In your case, you have 2^k inside your parentheses, and 2 before them. We already know that 2^k means “raise 2 to the power k”, or more simply, “multiply 2 by itself k times”.

    Also, there’s a convention – an agreed-on meaning – that putting two maths terms side by side (or “juxtaposing” them) – means that we should multiply them. So that an expression like abcde is a shortcut for writing a x b x c x d x e. But we have to be careful with terms that include numbers – there’s no point writing “22” to mean “2 x 2”, since we could confuse that with the usual decimal meaning of “22”, that is “twenty-two”. Still, we can write either “2 x 2” or “2 (2)” to keep them separate but adjacent (side by side).

    I’m guessing that you’ve heard of a rule called “BOMDAS” or something similar? This is just another maths convention that helps us ensure that whatever we write down in mathematical language has only one possible meaning – it helps us avoid confusion. For example, when you see the expression 2 + 5 x 6, how do you know whether you should:

    • add 5 to 2, then multiply the result by 6,
    • multiply 5 by 6, then add 2 to the result?

    The BOMDAS rule answers this question, by giving the order in which we should perform the different operations. Those operations, in order, are:

    1. Brackets [ … ], braces { …} and parentheses ( … )
    2. Of, as we find in expressions like “three-fifths of five-eighths of the total”, or “20 percent of Jack’s wages”
    3. Multiplication
    4. Division
    5. Addition
    6. Subtraction

    Applying this rule, we see that we have to multiply before adding, so that:

    2 + 5 x 6 = 2 + 30 = 32.

    In your case, we use the parentheses to keep the first 2 separate from the second 2, which represents the base 2 that we raise to the exponent k to get 2^k. But the RHS still means that we need to multiply 2 by 2^k.

    Now that we understand what each side of the equation means, how so we know it’s true? We use the first (secret) super-power of mathematics: our incredible observation skills! 😉 We notice that on the LHS we have the same term, “2^k”, repeated and added to itself. That’s the same as two lots of “2^k”, which we can write as “2 of 2^k.

    But what do we do with the word “of” in maths? In the expression “20 percent of Jack’s wages”, we can replace “of” by “multiply” or “x”. Supposing Jack’s wages are $300, then 20% of his ages is 20% x $300 = (20 / 100) x $300 = 0.2 x $300 = $60. And the expression “three-fifths of five-eighths of the total” just means “(3 / 5) x (5 /8) x (the total)”.

    So “2 of 2^k” is exactly the same as “2 x 2^k”, and that’s also the value of the RHS.

    Now, I’ve taken my time explaining this stuff. You’ll actually need to spend a lot more time than I did just now, to make sure you’ve got it. The only reason I understand any of this is because I have put in the time to think about it.

    One last thing: most of the trouble people have with maths is because nobody ever told them this: A maths class is a language class! As ever, if anything here is still unclear, feel free to ask more questions -and good luck with your language classes. 😉

  2. Let us assume [math]2^k[/math] as a.

    Now, the equation becomes a + a which is nothing but 2a.

    Let us put back the initial values for a= [math]2^k[/math]

    thus, [math]2^k[/math] + [math] 2^k [/math] = 2 ([math]2^k[/math])

    Happy Learning!!!

  3. Perhaps the sum of exponential functions leads you to believe that they are special and cannot be treated as regular quantities. But, if we choose to make the substitution [math]u = 2^x[/math], we see that

    [math]2^x + 2^x[/math]

    [math]= u + u[/math]

    [math]= 2u[/math]

    [math]= 2 \cdot 2^x[/math]

    This is not necessary, however; it should be intuitive that, by the way we define addition and multiplication, adding something to itself is equivalent to multiplying that thing by [math]2[/math]!

  4. [math]2^k + 2^k = 2(2^k)[/math]

    let [math]a = 2^k[/math]

    [math]a + a = 2(2^k)[/math]

    [math]2a = 2(2^k)[/math]

    [math]2(2^k) = 2(2^k)[/math] substitute

  5. Simple. Using the left hand side (LHS) of the equation…

    [math]LHS = 2^k+2^k[/math]

    From here, factorize it.

    [math]LHS = 2(2^k) = RHS[/math]

    Note: RHS stands for Right Hand Side

    You can also do that for RHS.

    [math]RHS = 2(2^k)[/math]

    From the equation, we know that from the equation there is two of [math]2^k[/math], so it make sense to split them up into this…

    [math]RHS = 2^k(1+1) = 2^k 2^k = LHS[/math]

    See? Since LHS=RHS, you have proven the statement, [math]2^k+2^k = 2(2^k)[/math].

  6. Let [math]x[/math] be any real number (such as [math]2^k[/math] for real [math]k[/math]), then [math]2x = (1 + 1)x = x + x[/math] by elementary algebra.

  7. May I ask you, what is multiplication exactly?

    Multiplication is adding a number to itself a specific number of times.

    Here are some examples:

    1 + 1 = 2(1) = 2

    2 + 2 = 2(2) = 4

    3 +3 = 2(3) = 6

    4 + 4 = 2(4) = 8

    5 + 5 + 5 = 3(5) = 15


    2^k + 2^k = 2(2^k)

  8. it can be shown easily by using mathematical induction but let me use simple logic and conclude the same result.

    first assuming k ≠ -∞ then divide both by 2^k

    left side : 2^k / 2^ k + 2^k / 2^ k which leads to 1+1 = 2(1)

    Right side : 2 * (2^k / 2^k ) = 2(1)

    Since both sides are equal then 2^k + 2^k = 2 * 2^k.

  9. If you have one [math]2^k[/math], and you add another [math]2^k[/math], then all together you have two [math]2^k[/math]’s.

  10. Suppose you have [math]a \in \mathbb{R}[/math]

    [math]a + a = 2(a) [/math]

    Here we fix k, and let [math]a^{k} = b [/math] then

    [math]a^{k} + a^{k} = b+b = 2(b) [/math]

    so for any a here if we fix k this a fact that simply adding something here is equivalent to multiplication its repeated addition.

  11. There are few ways to look at this question.

    But the easiest way is to make 2^k into x.

    Let 2^k be x, x+x = 2x

    from here you know x+x is probably equal to 2x.

    2x =2x (shown, L.H.S = R.H.S)

    Sub x into 2^k, 2(2^k) = 2(2^k).

    It would be clear to understand the question by using simple algebra expression which simplified the question. If there is any doubt, feel free to ask me or drop me a message.

  12. Think of it like this:

    1(2^k) + 1(2^k) = 2(2^k), because any number is the same as 1 * itself.
    Now factor 2^k out to get this:
    2^k(1 + 1) = 2(2^k)
    Which simplifies to
    2(2^k) = 2(2^k)
    So, both the expressions are equal, by using a simple proof.

  13. The simplest method possible

    2^k=(2^k) x 1= 1(2^k)

    substitute this in to the original equation


    factor out the 2^k





  14. Anything multiplied by 1 remain same so we can easily write above as :

    1 x (2k) + 1 x (2K)

    Now since 2K is common in both, we may write it as follows:

    2K (1 + 1) or (1+1)2K




  15. The only reason this doesn’t make perfect sense to you is that you are doing something very familiar using unfamiliar symbols. I’m not going to give you a technical explanation. I’m just going to show you how it makes sense.

    When you learned how to multiply, you learned that…

    4 + 4 = 2 x 4 … or as you are writing it now… 4 + 4 = 2(4). Similarly…

    4 + 4 + 4 = 3(4)

    4 + 4 + 4 + 4 + 4 + 4 = 6(4)

    Now you know that … x + x = 2(x) , or as we write it more simply … 2x

    x + x + x = 3(x) or 3x

    x + x + x + x + x + x = 6(x) or 6x … and so on.

    As long as you are adding the same thing, you can change an addition into a multiplication. So …

    (2^k) + (2^k) = 2(2^k)

    (2^k) + (2^k) + (2^k) = 3(2^k) and so on…

  16. Well quiet simple (preschool level really):

    [math]\Huge{A: 1+1=2}[/math]

    Do we all agree on this?

    [math]\Huge{B: a+b=c \iff \overbrace{(a+b) \times d}^{=ad+bd}=c \times d}[/math]

    Also quiet clear?

    Now try to combine [math]\Huge{\text{A and B} \Huge \color{red}{\text{ logical}}}[/math] (hint [math]d=2^k[/math]) and voila, this is how it works.

  17. It’s pretty simple, let’s see. Usually you know you add two or more things that are equal. Like x+x=2x. In this case you are adding two thing leaving it indicated.

  18. [math]2^k+2^k=2(2^k)[/math]

    Take LHS, Take [math]2^k[/math] as common factor, we get,


    We know that [math]1+1=2[/math], so, the equation would reduce to,

    [math]2^k(2)[/math] or [math]2(2^k)[/math]

    I hope my answer was helpful. If you've any questions regarding Mathematics, Physics or Computers, ask me.

    Good day my math friend!!!

  19. divide both sides by [math]2^k[/math] and you get the 1 + 1 = 2. The only case when you can not do the divide would be at negative inifinity when the denominator would be 0

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