Which numbers is bigger $80^{105}$ or $28^{140}$

Question

Which numbers is bigger $80^{105}$ or $28^{140}$

My try follows

$80^{105}$ = $(80^{3})^{35}$=$(512000)^{35}$

$28^{140}$=$(28^{4})^{35}$=$ (614656)^{35}$

So $28^{140}$ is bigger

Is there another idea for this problem involving relations between $80$ and $81$ and $27$ and $28$ since $81$ and $27$ have common base $3$

Thank you for your help

Answer 1

$$28^{140}>27^{140}=3^{420}=81^{105}>80^{105}$$

Answer 2

Take logarithms, since $ \log{x} $ is monotonic. You can even use base 10.

Then $ \log_{10}{80^{105}} = 105 \log_{10}{80} \approx 200 $. Meanwhile, $ \log_{10}{28^{140}} = 140 \log_{10}{28} \approx 203 $. Hence $ 28^{140} $ is bigger.

(Edit: This isn't as clever as the previous answer since you need a computer for it, but it can be extended to powers that are difficult to compute).

Answer 3

As $(105,140)=35$

we can check for $80^3,28^4$

$80^3=(5\cdot2^4)^3=5^32^{12},28^4=(7\cdot2^2)^4=7^4\cdot2^8$

$80^3>=<28^4\iff5^32^{12}>=<7^4\cdot2^8\iff5^32^4>=<7^4$

Now $7^4=2401,5^32^4=2\cdot10^3=2000$

Answer 4

kingW3's answer, which invokes only the inequalities $27\lt28$ and $80\lt81$, can hardly be beat, but here's another, lengthier, approach, invoking three inequalities ($48\lt49$, $8\lt9$, and $125\lt128$):

$$\begin{align} 48\lt49&\implies2^4\cdot3\lt7^2\\ &\implies2^8\cdot3^2\lt7^4\\ &\implies2^8\cdot2^3\lt7^4\quad(\text{since }8\lt9)\\ &\implies2^4\cdot2^7\lt7^4\\ &\implies2^4\cdot5^3\lt7^4\quad(\text{since }125\lt128)\\ &\implies(2^4\cdot5)^3\lt(2^2\cdot7)^4\\ &\implies80^3\lt28^4\\ &\implies80^{105}\lt28^{140} \end{align}$$