Solve this quickly!

$7^{10}=49^{5}>40^5=4^5\times10^5>10^7$ since $4^5=1024$

take log , log 10^7 = 7 log 10 = 7 x 1 = 7 and log7^10 = 10 log 7 = 10 x 0.84 = 8.45

so B > A

$7^{10} > 49^{5} > 32^{5} = 2^{25} = 2^{7}×2^{18} = 2^{7}×32^{3.6} > 2^{7}×25^{3.5} = 2^{7}×5^{7} = 10^{7}$

Consider the function, y=x^{1/x}. Evaluate dy/dx and it turns out to be x^{(1/x)-2} (1-ln x), which when equated to zero gives x=e. Note that dy/dx >0 for x <e, dy/dx=0 for x=e and dy/dx <0 for x>e. Therefore the function y=x^{1/x} is a decreasing function when x>e. Hence, for x1>x2>e, x2^{1/x2} > x1^{1/x1}.

Since 10>7>e, 7^(1/7) > 10^(1/10). Raisiing the power to 70 on both sides, 7^10 > 10^7

7x7x7x7x7x7x7x7x7x7=282475249>10x10x10x10x10x10x10=1000000

Simply 10^7 is (0.01)^10 which is less than 7^10 :) :)

282475249>10000000 probably B is correct

For $10^7$ there will be seven zeroes after one

$10^7 = 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 = 10000000$

Now,

$7^{10} = 49^5$ or $282475249$

Therefore $7^{10}$ is larger.

if i take log ,then log 10^7=7 log 10=7 and log 7^10=10 log 7=8.4509804(calculator). so,7^10 is bigger

If you know that $7^3 = 343$ , then you can estimate that $7^6 = 343^2$ is a little greater than $100000 = 10^5$ .

With that in mind, base 10 logarithms tell us that $\log(7^6)\ge\log(10^5)\;\Rightarrow\;6\log7\ge5\;\Rightarrow\;\log7\ge\frac56$ .

Thus, $\log(7^{10}) = 10\log7 \ge 10\cdot\frac56 = \frac{25}3$ .

Since $\log(10^7) = 7 < \frac{25}3$ , then $\log(10^7) < \log(7^{10})$ , which means $7^{10} > 10^7$ .

I'm wondering why there isn't anybody who used log to solve this: 10log7 =10x0.9..... >0.9 > 7log10=7

Using logarithms, it is easy to see that 7^10 is the larger number. But, actually, common sense with numbers would allow you to intuit correctly that 7^10 is the larger number.

= $7^{10} ... 10^{7}$ , = $7^{3}$ ... $10$ , $= 343 > 10$ . So, $7^{10} > 10^{7}$

Since power is bigger number of significant figures is more thus 7^10 is bigger than 10^7.

7^10>6^10=6^7×6^3 6^3=2^3×3^3>2^3×2^4=>2^7 Therefore, 7^10>6^10>6^7×2^7=12^7>10^7 Hence proved

I cheated and used the fact that I have the first ten powers of 2 memorized:

4<7<8, so (4^10)<(7^10)<(8^10)

4^10=(2^10)^2 or roughly 10^6

8^10=(2^10)^3 or roughly 10^9

7 is closer to 8, so 7^10 is going to be closer to 10^9 which is larger than 10^7

10 exponent 7 = 100M 7 exponent 10 =70B therefore 7 exponent 10 is larger than 10 exponent 7.

simply count the digit roughly... just multiply and count the digit and you will find which one is bigger

But how to generalize this concept for any two numbers a & b, where as a > b ? Any friend can clear ?

f(x) = ln(x)/x (x>=1); lim as x --> +infinity of f(x) = 0 so ln(x1)/x1 > ln(x2)/x2 with x1 < x2 (ln7)/7 > (ln10)/10 => 10ln7 > 7ln10 => ln(7^10) > ln(10^7) => 7^10 > 10^7

Subtitution with ~30% so it goes like this ::: 7^3 = 343 /// 10^2 = 100 so 10^2.1 should be just around 100. Which larger ? You dont even have using calculator.. Prove yourself :)

Check this one...... Log 10^7 < log 7^10...... 7log 10 < 10 log 7 ...... 7x1 < 10x0.84

10^7 ? 7^10

1 ? .7^7*7^3

1? (.7)*(3.43)^3

70% of 3 itself is greater than 1

So 1< (.7)*(3.43)^3

Hence 7^10 is bigger.

We know that 7^8 > 10^7, and therefore 7^10 >> 10^7. Thats it

every time you multiply with 7 you add number that is mean the final number will consist of 10 numbers

2^3=8,3^2=9 and 3^4=81,4^3=64 when the power is greater than base product is greater.So 7^10 is greater than 10^7

Let's put "?" for our relation.

10⁷ ? 7¹⁰ log(10⁷) ? log(7¹⁰) [apply log, which is increasing, so the "?" sign stays the same] 7.log(10) ? 10.log(7) [log property] 7 ? 10.log(7) [log(10=1)]

Now, log(7)>1, so, 10 log(7)>10, therefore 7 < 10.log(7) which means that "?" = "<" and then

10⁷ < 7¹⁰

(7/10)^7=0.7^7, this value may be less than one but after a point you will have some number. remaining 7^3 =21. if you multiply both ,the answer must be larger than one.so answer wold be B

10^7=(7x1.43)^7 so divide both by 7^7 so 1.43^10 versus7^3 is no brainer 7^3 is much larger

Think of these spatially!