2
$\begingroup$

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

$\endgroup$

4 Answers 4

11
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ Nice answer you wrote there. $\endgroup$
    – Iti Shree
    May 1, 2017 at 14:57
  • $\begingroup$ @kingW3 vElegant answer, thank you very much +1 $\endgroup$
    – Medo
    May 1, 2017 at 15:11
2
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ @BarryCipra, Thanks $\endgroup$ May 1, 2017 at 16:19
1
$\begingroup$

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 $125\lt128$):

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.