Exponent Challenge

Relevant wiki: Rules of Exponents

I have decided to post a more detailed solution which is meant to explain how we got to the answer we did to anybody who is confused by, or has no clue how people came up with answers that just say something like: $2^{987} - 2^{986} = 2^{986}\cdot (2 - 1) = 2^{986}.$

If the solution I just used as an example makes sense to you, then this post is not for you, but if you are like many of the people who have asked me how on earth you get from $2^{987} - 2^{986}$ to $2^{986}\cdot (2-1)$ then please read on.

We will use what's known as factoring to solve this problem. First, lets use the rules of exponents to create "like terms" which we can factor out. We know as a rule of exponents that $a^b \cdot a^c = a^{b+c}.$ so using this rule I'm going to split up 2^987 into this: $2^{986} \cdot 2^{1}.$

Now the problem reads as follows: $\left(2^{286} \cdot 2^1 \right) - 2^{986}$ which is simply $\left(2^{986} \cdot 2\right) - 2^{986}.$

At this point, you may already understand how to solve this; if not, I'll solve this 2 different ways. The first is the commonly accepted way and the second will just be for anybody who needs an even simpler approach.

The first way we factor out the common term: $2^{986}.$ That means you essentially divide the entire thing by $2^{986}$ and then multiply it by $2^{986}$ again so that the net effect is no change. Now you have: $2^{986} \cdot 1 = 2^{986}.$

The second way is just an easier way to show the solution for anybody who didn't quite grasp the first solution. One really cool concept of a variable is that you can set it equal to anything you want it to be, in this case a big exponent like $2^{986}$ is what makes this problem hard to solve for some people. Let's simply replace all the $2^{986}$ we see with a variable, $n.$ Then, we have $(n \cdot 2)-n = 2n-n = n = 2^{986}.$

So by replacing something that looked like a rather hard to grasp number with a variable, we were able to more easily reduce the problem. Please note in special cases, this method may not quite work. This was just for clarity and to show people a new way of looking at things that may be easier to understand. Rest assured that if you do encounter one of these special cases, you will likely understand math well enough to not need this approach.

Hopefully this solution was a little more clear than the other ones and helped somebody out there to understand how we got the answer we did.

Relevant wiki: Rules of Exponents

$\large{\begin{aligned} & 2^{987} - 2^{986} \\ &= 2^{986} (2^{1} - 1) \\ &= 2^{986} (2-1) \\ &= 2^{986}.\end{aligned}}$

2^987 - 2^986 = 2^(986 + 1) - 2^986 = 2^986 * 2 - 2^986 = 2^986(2 - 1) = 2^986

${ 2 }^{ 987 }-{ 2 }^{ 986 }$

= ${ 2 }^{ 986 }(2-1)$

= ${ 2 }^{ 986 }(1)$

= $\boxed{{ 2 }^{ 986 }}$

2 to the power 986

2^987 - 2^986 = 2^986(2 - 1) = 2^986 * 1 = 2^986

2^986x(2-1)=2^986

2^987-2^986=2^986(2-1)=2^986 ans

simple logic - a number is double of the number so the number will be the number itself.

We can express these as 2^286(2-1) that is 2 ^286 Ans

Hit and Trial Method 2^6 - 2^5 = 64-32= 32 [2^5] 2^7 - 2^6 = 128-64= 64 [2^6] Therefore, 2^987 - 2^986= 2^986

Let x = 986, so the problem is 2^(x+1) - 2^x, so problem is 2^x(2^1 - 2^0) = 2^x(2-1) = 2^x QED

The values of the powers of 2 from 2^0 upwards are 1,2,4,8,16,32,64,128, 256 etc. Take any pair, 2^N and 2^(N-1), e.g. (128 and 64) and you will notice the difference between them (128-64) is always the smaller number 2^(N-1).

$2^{287}$ - $2^{286}$ - Given

$2 \times$ $2^{286}$ - $2^{286}$ -Factor out 2

$2^{286}$ - Subtract like terms

2^986 Let x be 2^986 So 2^987 can be factor as 2(2^986) Since x=2^986 So 2x-x=x x=2^986

Each multiplication in an exponent 2^x is a multiplication n(2), which is the same as n + n.

2^2 = 2(2) = 2 + 2 = 4

2^3 = 4(2) = 4 + 4 = 8

2^4 = 8(2) = 8 + 8 = 16

...

2^987 = 2^986(2) = 2^986 + 2^986

2^987 = 2^986 + 2^986

2^987 - 2^986 = 2^986

So,

(2^n+1) - (2^n) = 2^n

$2^{987}$ is $2^{986} \times 2$ , so we're basically halving the double.

Do you agree that 2^987is double the number of 2^986.Then any number which is subtracted from the double of its number is going to be the same number. So 2^987-2^986 is 2^986

2^987 - 2^986 = 2^986.(2^1 - 1) = 2^986.(2 - 1) = 2^986.1 = 2^986

Exponent Theorem 1.0 (I am not sure if someone has claimed or proved this) can solve this.

- The I'm-certain-not-everyone-will-agree-with "proof" can be accessed through https://brilliant.org/discussions/thread/exponent-theorem-10.

The theorem states that $b^n - b^{n-1} = b^{n-1} (b-1)$ .

By substitution,

$2^{987} - 2^{987-1} = 2^{987-1} (2-1)$

$2^{987} - 2^{986} = 2^{986} \times 1$

By multiplication identity property,

$\boxed{2^{987} - 2^{986} = 2^{986}}$

2^987 - 2^986 = 2^986(2 - 1) = 2^986 x1 = 2^986

=(2^n)-(2^n-1) =(2^n)-(2^n 2^-1) =(2^n)-(2^n)/2 =(2(2^n)-(2^n))/2 =(2^n)/2 =(2^n)2^-1 =2^(n-1) here n=987 hence 2^(987-1)=2^986

for example, 2^4 - 2^3 = 2^3 2^7-2^6=128-64=2^6 in the same way 2^987-2^986=2^986