Comparison

The two numbers $80$ and $28$ are close to a number with a raised power, so :

$80^{105} < 81^{105}=\left( 3^{4}\right) ^{105}=3^{420}$

$28^{140} > 27^{140}=\left( 3^{3}\right) ^{140}=3^{420}$

therefore :

$80^{105} < 3^{420} < 28^{140}$

\fbox{\$80^{105} < 28^{140}\$}

Find the 35th root of them then we will get $80^3$ and $28^4$ As $80^3 = 512,000;\:28^4 = 614,656\Rightarrow 80^{105}<28^{140}$

Just Find the number of digits in each number.

Computing the ratio

$R = 80^{105} : 28^{140} = 4^{210}\times 5^{105} : 4^{140}\times 7^{140}$

$\displaystyle = 4^{70}\times 5^{105} : 7^{140}$

$\displaystyle = (4^2 \times 5^3)^{35} : (7^4)^{35}$

$\displaystyle = (2000)^{35} : (2401)^{35}$

$\displaystyle \Rightarrow R < 1$

$\displaystyle \Rightarrow \boxed{80^{105} < 28^{140}}$

Let a=80^105 Applying log to the base 10 on both sides we get loga= 150log80 Then after multiplication we get a=10^199.82. Now b=28^140 same as above we get b=10^202.60 a<b

Use Log Base Transform: 80^105 is equal to 28^138.0805995 28^138.0805995< 28^140

much simpler solution balancing the outer exponents: (28^4)^35 > (80^3)^35