Which of These Gargantuan Numbers Is Bigger?

Well, $4^{5000} \times 4^{5000}$ is equal to $4^{10000},$ which is equal to $2^{20000}.$

On the other hand, $2^{19999}+2^{19999}$ can be expressed as $2^{19999}*2$ or $2^{20000}$ , thus the answer is They are equal.

• Just Use the Product Rule and place the result in the Base $2$

$A = 4^{5000} \times 4^{5000}$ = $4^{10000}$ = $2^{2 \cdot 10000}$ = $2^{20000}$

• Just put $2^{19999}$ on Evidence

$B = 2^{19999} + 2^{19999}$

$B = (2^{19999})(2)$ $\Rightarrow$ $2^{19999 + 1} = 2^{20000}$

So $A$ and $B$ Are Equal

Given that, $A=4^{5000} . 4^{5000}$ and $B=2^{19999}+2^{19999}$

Now, $A=4^{5000} . 4^{5000} = (4\times 4)^{5000}=(16)^{5000} = (2^4)^{5000} = (2)^{4\times 5000}=\boxed{2^{20000}}$ ....(i)

Also, $B=2^{19999}+2^{19999}=2\times 2^{19999} = 2^{1+19999} = \boxed{2^{20000}}$ ....(ii)

From (i) and (ii), we see that A=B

So, the answer is Both the values are equal.

Solving A: we know that $4 = 2^2$ , so we have this value to A: $(2^2)^{5000} * (2^2)^{5000}$ . Using this rule: $(x^m)^n = x^{m*n}$ in each terms and then this one: $x^m*x^n = x^{m+n}$ , we get this value to A: $A = 2^{20000}$ .

Solving B: when we have $3 + 3$ , we can express this value as $3*(1 + 1)$ . We just need to do exactly the same here: $B = 2^{19999} + 2^{19999} \Rightarrow B = 2^{19999}*(1 + 1) \Rightarrow B = 2^{19999}*2$ and then we make as above and we have $B = 2^{20000}$ .

Finally, $A = B$

First we simplify these two terms first

$A = 4^{5000}\cdot4^{5000}$

$=4^{10000}$

$=(2^2)^{10000}$

$=2^{20000}$

While $B = 2^{19999}+2^{19999}$ is similar to $2^{19999}\cdot2$

$2^{19999}\cdot2 = 2^{20000}$

Therefore, $A$ is equal to $B$

A=4^10000 B=2(2^19999)=2^20000 They are equal

A = $4^{5000}$ * $4^{5000}$ = $4^{5000+5000}$ = $4^{10000}$ = $(2^{2})^{10000}$ = $2^{20000}$

B = $2^{19999}$ + $2^{19999}$ = $2$ * $2^{19999}$ = $2^{19999+1}$ = $2^{20000}$

A=B

2^19999 + 2^19999 = 2X2^19999 = 2^20000 4^5000 X 4^5000 = 4^5000+5000 = 4^10000 = (2^2)^10000 = 2^20000 Hence both the values are equal

with 2 as base;exponents are equal..so A=B

2^19999+2^19999=2^20000 4^5000+4^5000=4^10000=2^20000

$4^{5000} = (2^{2})^{5000} = 2^{10000}$

$4^{5000} \times 4^{5000} = 2^{10000} \times 2^{10000} = 2^{10000+10000} = 2^{20000}$

$2^{19999} + 2^{19999} = 2 \times (2^{19999}) = 2^{20000}$

It's a Nice tricky Problem :) Find A = 4^5000+5000 that means 4^10000 = 40000

Find B = 2^19999+1 = 2^20000 = 40000

so they are equal

$A= 4^{5000}.4^{5000}= 4^{10000}$ and $B= 2^{19999}+2^{19999}= 2.2^{19999}= 2^{20000}= {(2^2)}^{10000}= 4^{10000}$

$A = 4^{5000}.4^{5000} = 2^{10000}.2^{10000} = 2^{10000 + 10000} = 2^{20000}$

$B = 2^{19999} + 2^{19999} = 2^{19999}(1 + 1) = 2^{19999}.(2) = 2^{19999}.2^{1} = 2^{19999+1} = 2^{20000}$

Thus, $\boxed{A = B}$

A=2^2^5000 X 2^2^5000 =2^20000

B=2X2^19999 =2^20000

hence A=B

A = 4^5000 . 4^5000 = 2^10000 . 2^10000 = 2^20000

B = 2^19999 + 2^19999 = 2^19999 (1+1) = 2^19999 . 2 = 2^20000

:. A = B

They are they same.

$A = 4^{5000} \cdot 4^{5000} = 4^{10000} = (2^{2})^{10000} = 2^{20000}$

$B = 2 ^{19999} + 2^{19999} = 2 \cdot 2^{19999} = 2^{20000}$

SIMPLE again : follow that laws of indices apply only to products , for summation , take 2 e19,999 common and proceed., so both are equal.

$A=4^{5000}\cdot4^{5000}=(2^2)^{5000}\cdot(2^2)^{5000}=2^{10000}\cdot2^{10000}=2^{20000}$

and

$B=2^{19999}+2^{19999}=2\cdot2^{19999}=2^{20000}$

Therefore, they are equal.

I liked this question

2^20,000=x. 2^19,9999=x/2. x/2+x/2= x. Thus they are equal.

4^5000.4^5000=2^10000*2=2^20000 and 2^19999+2^19999 is also equals to 2^20000becouse we take 2^19999 common fro, both the terms and which becomes 2^19999(1+1)=2^19999(2)=2^19999+1=2^20000.

A=4^5000 X 4^5000= 4^10000,

B=2^19999 + 2^19999 multiply and divide by 2 we get B=2((2^19999)+(2^19999))/2, B=(2^20000 + 2^20000)/2, B=(4^10000 + 4^10000)/2, B=4^10000(1+1)/2, B=4^10000 (2)/2=4^10000,

A=B.

I think you need no law, states that 5000=5 and 19999 approx = 19 try to solve it that way 4^5.4^5 and 2^19+2^19 you will get the same answer. Note : not sure if this answer could be wrong in some cases but it worked for that problem :)

$A=\left ( 2^{2} \right )^{5000}\times \left ( 2^{2} \right )^{5000}=2^{20000}$ $B=2^{19999}\times \left ( 1+1 \right )=2^{20000}$ $=>A=B$

4^5000 * 4^5000 = (4^5000)^2= 4^10000= FINAL: 2^20000. 2^19999+2^19999= 2(2^19999)= (2^1)*(2^19999)= FINAL: 2^20000 So there are equal

let start with a simple solution 2^2+2^2=4+4=8, means 2^3 so 2^19999+2^19999=2^19999+1=2^20000 and 4^500 4 500=2^20000 (:-4^500=2^500.2^500 = 2^10000)

Let's see A = 4^5000 * 4^5000 = 4^(5000+5000) = 4^10000 = (2^2)^10000 = 2^20000. On the other hand, B = 2^19999 + 2^19999 = 2* 2^19999 = 2^20000. So that, A=B

4^5000*4^5000=4^10000=2^20000 and also 2^19999+2^19999=2^20000

4^{5000} * 4^{5000} = 2^{10000} * 2^{10000} = 2^{20000} = 2^{19999} * 2 = 2^{19999} + 2^{19999}

A= 4^(5000+5000)= 4^10000 =2^20000 B= 2^19999 * 2 = 2^20000 Hence,they are equal.

A=4^{5000}.4^{5000} =2^{10000}.2^{10000} =2^{20000}

B=2.2^{19999} =2^{20000} Hence, A=B.

4^5000 . 4^5000 = 4^10000 = 2^20000 2^19999+2^19999= 2*2^19999=2^20000

A=(4^5000).(4^5000)=(4^10000)=(2^20000). Again, B=(2^19999)+(2^19999)=2.2^19999=2^200000 So, A=B

4^{5000} . 4^{5000} = 4^{10000} = 2^{20000} = 2^{19999} + 2^{19999}

A = 4^10.000; B = 2^20.000 = (2^2)^10.000 = 4^10.000 =A

A = (2^2)^(5000X2) = 2^(2x10000) = 2^20000. And B = 2 x 2^19999 = 2^(19999+1) = 2^20000. So A and B are equals.

4^5000 =2^2(5000)=2^10000 =>2^10000 X 2^10000= 2^20000

2^19999 +2^19999 = 2^19999(1 +1) = 2^19999 . 2 = 2^20000

a=2^20'000 b=2^19999(1+1)=> 2^20000 so a=b

A = 2^{10000} . 2^{10000} = 2^{20000} and B = 2^{19999} . (1+1) = 2^{19999}.2 = 2^{20000} Therefore, A=B

A=4^5000 4^5000=2^10000 2^10000=2^20000

B=2^19999+2^19999=2^(19999+19999)=2^20000

A can be written as follows:

$2^{10000} \times 2^{10000} = 2^{20000}$

B can be written as follows:

$2^{19999} + 2^{19999} = 2 \times 2^{19999}$

$2^{19999} = \frac {2^{20000}}{2}$

Therefore A and B are equal.

here A=4^5000.4^5000 =A=2^2^5000 2^2^5000 =A=2^10000 2^10000 A=2^20000

WHEREAS B=2^19999+2^19999 B=2*2^19999 B=2^(19999+1) B=2^20000 SO A=B

*HERE ^=TO THE POWER *=X(MULTIPLIED)

You can verify that $a^{b}$ $\times$ $a^{c}$ = $a^{(b+c)}$ For example, $3^{2}$ $\times$ $3^{3}$ = 9 $\times$ 27 = 243 This can also be computed as, $3^{2}$ $\times$ $3^{3}$ = $3^{(2+3)}$ = $3^{5}$ = 243 Using this principle, $4^{5000}$ $\times$ $4^{5000}$ = $4^{(5000+5000)}$ = $4^{10000}$

Also, $a^{b^{c}}$ = $a^{b \times c}$ . Since, 4 = $2^{2}$ , we can write $4^{10000}$ as $2^{2^{10000}}$ = $2^{(2 \times 10000)}$ = $2^{20000}$

You can verify that $a \times a^{n}$ = $a^{(n+1)}$ . For example, $2 \times 2^{3}$ = $2 \times 8$ = 16. This can also be computed as, $2 \times 2^{3}$ = $2^{1} \times 2^{3}$ = $2^{(1+3)}$ = $2^{4}$ = 16. Using this principle, $2^{19999}$ + $2^{19999}$ = $2 \times 2^{19999}$ = $2^{(1+19999)}$ = $2^{20000}$

Take common as 2^19999 from B v get 2^20000 From A, as bases are equal ,it will become 4^10000=2^20000