Can't see the 3 for the forest

3^12345 - 3^12344 = 3*(3^12344) - 3^12344.

Taking 3^12344 as a common factor,

3^12344 (3-1) = 2 (3^12344)

If one is looking for a more logical and visual explanation, you could simply say that 3^12345 is 3 groupings of 3^12344, and that by subtracting 3^12344, you are merely taking away one of the 3 groupings, meaning what is left behind are 2 of these grouping, therefore the answer is 2 * 3^12344

$3^{12345}$ is going to be 3 times the size of $3^{12344}$ . If we say $x = 3^{12344}$ , we can then write the initial equation as $3x - x$ , which, of course, is going to be $2x$ . So the answer is $2 * 3^{12344}$ .

I did this to figure it out: 3^3 - 3^2 = 27-9 = 18 = 2 9 = 2 3^2 thus by analogy, the answer must be (and of course is):
2 * 3^12344

$3^{12345}-3^{12344}=$ $3(3^{12344})-3^{12344}=$ $3^{12344}+3^{12344}+3^{12344}-3^{12344}=$ $\boxed{3^{12344}\times2}$

Taking $12344$ as $x$ , let's rewrite the expression as, $3^{x+1}-3^{x}$ $=3.3^{x}-3^{x}$ $=2.3^{x}$ $=\boxed{2\times3^{12344}}$

Let y = 3^12345 - 3^12344 so y is the answer we want to get.

We want to do something to eliminate one of the 3^..., so what about dividing both sides by 3^12344?

So we now have: (3^12345)/(3^12344) - (3^12344)/(3^12344) = y / (3^12344)

The first term, can be written as 3^(12345-12344), right? That is equal to 3^1 = 3.

For the second term, the upper and lower sides of the fractions are equal which easily shows the fraction is equal to ONE, and that what we wanted (to see less 3^... in the left side).

Now, you have: 3 - 1 = y / (3^12344) 2 = y / (3^12344)

Here, you can cross-multiply to make "y" the subject, which will result in.. y = 2 * 3^12344

Yay!! It's actually one of the choices, and it's an easier way to represent the answer. :D

Let $3^{12345}-3^{12344}=x$

Dividing both sides by $3^{12344}$ , we get

$\frac{3^{12345}}{3^{12344}} - \frac{3^{12344}}{3^{12344}} = \frac{x}{3^{12344}}$

$3-1=\frac{x}{3^{12344}}$

$x=2×3^{12344}$

3^12345-3^12344=3^12344(3-1)
thus, the answer is 2 * (3^12344)

let 3^12345 =3 X

let 3^12344 =X

3X-X=2X

X = 3^12344

→ 2X = 2*3^12344

3^12345 - 3^12344 = 3(3^12344) - 3^12344. By taking 3^12344 as common we get 3^12344(3-1) = 2*(3^12344).

They had this same question yesterday

We have 3^12345-3^12344

We can take 3^12344 as common

Now, we get

3^12344 (3-1)

2×(3^12344)

My exponent theorem @ https://brilliant.org/discussions/thread/exponent-theorem-10 states that $b^n - b^{n-1} = b^{n-1} (b-1)$ .

By substitution,

$3^{12345} - 3^{12345-1} = 3^{12345-1} (3-1)$

$3^{12345} - 3^{12344} = 3^{12344} \times 2$

$\boxed{3^{12345} - 3^{12344} = 2 \times 3^{1234}}$

3raise to 12344(3-1)=2x3raise to 12344

3^12345 - 3^12344 = 3^(12344+1) - 3^12344 = (3 ^ 1 - 1) x 3^12344 = 2 x 3^12344

3^12345-3^12344=3^12344 (3^1-1)=2*3^12344

The first one (3^12345) is 3 times bigger than the another one (3^12344).

So if u Subtract (3^12344) just one time from (3^12345) than u get not more 3 x (3^12344), but just 2 x (3^12344).

3^12344(3-1)=3^12344*2

Which means we took 3^12344 as coomon term then remaining equal to 3 minus 1 which equal 2*3^12344

I don't know, I just guessed it

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Just simplify the equation into 3^2 - 3^1 = 9 - 3 = 6 ... where 6 actually equals 2 * 3^1 ..

Repeat using 3^12344 instead of 3^1 :)

3^{n+1} - 3^n=3^n (3-1)=2 \cdot 3^n \n=12344 \ \therefore 3^{12345} - 3^{12344}= 2 \cdot 3^{12344}

$3^{12345}-3^{12344}$

Factor out $2^{12344}$ .

$3^{12344} ( \frac{3^{12345}}{3^{12344}}-\frac{3^{12344}}{3^{12344}})\\=3^{12344}(3-1)\\=3^{12344}(2)$

$=\boxed{2\times3^{12344}}$

a^n - a^(n-1) = a^n - a^n.a^-1 = a^n . (1 - 1/a ) = a^n . (a-1)/a = (a^n)/a . (a-1) = (a^(n-1)) (a-1) If a=3 & n=12345 3^12345 – 3^12344 = 2 . 3^12344

LIKE 3^3 - 3^2 27-9 =18 NOW 18 ,and we write that 2x3^2 same as 2x3^12344

3^12344 x3- 3^12344 =3^12344(3-1) =2X3^12344

Logically, if you look at them closely, none of the other three options are mathematically possible. 2x3^12344 is the only possible answer. Not a very mathematical solution maybe but the 'process of elimination' is easy here.

3^12345-3^12345=3^(12344+1)-3^12344=3 3^12344-3^12344 taking 3^12344 common....... 3^12344(3-1)=2 3^12344 (Ans.) .

3^12345-3^1234 =3^1234(3-1) =3^1234(2)

3^12345 -3^12344 = 3x3^12344 -1x3^12344) = (3-1)x3^12344= 2x3^12344

3^12344(3-1) = 2 3^12344 , ( 3^12344 3^1)-3^12344 = 3^12344(3-1) 3^12344 2 2 3^12344