Pawan karki

$56^5 < 125^5 = 5^{3\times 5} = 5^{15} < 5^{56}\quad \Rightarrow \boxed{5^{56} > 56^5}$

$5^{ 56 }=(5^{ 2 })^{ 28 }=(25^{ 2 })^{ 14 }>56^{ 5 }$

And this is true as:

• $25^{2}>56$

• $14>5$

$5^{56}>56^5\iff 5^{1/5}>56^{1/56}$ , which is true, since

$x^{1/x}$ is strictly decreasing in $(e,+\infty)$ , since

$(x^{1/x})'=-x^{\frac{1}{x}-2} (\log(x)-1)<0$ for $x>e$ .

if the two numbers are to be compared with base and power being the same numbers, the number with higher power is larger than the number with lower power. EXCEPTIONS: A. 2^3 and 3^2( here 3^2 is larger) B. 2^4 and 4^2 (here both are equal)

${ 5 }^{ 56 }$ in base 5 is 57 digits long. How many digits ${ 56 }^{ 5 }$ is going to be in base 5? Not that many.

$5^{56} = (5^{2})^{28} = 25^{28} = (5^{4})^{14}$ or $(25^{2})^{14} = 625^{14}$

since 625 > 56 and 14 > 5, therefore

$625^{14}$ is much greater than $56^{5}$

$\boxed{5^{56} > 56^{5}}$

5^56 = 25^28 = 125^14 > 56^5

Which is Greater?

A. 5^56 =

1.38777878 E+39 =

1387778780781445675529539585113525390625

or

B. 56^5 =

5.50731776 E+8 =

550731776

Final Answer: A is greater than B.