MHB Which is Greater: $\log_{10}12$ or $(\log_{10}5)^2+(\log_{10}7)^2$?

  • Thread starter Thread starter anemone
  • Start date Start date
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
State with reason which of these is greater?

$\log_{10}12$ or $(\log_{10}5)^2+(\log_{10}7)^2$?
 
Mathematics news on Phys.org
anemone said:
State with reason which of these is greater?

$\log_{10}12$ or $(\log_{10}5)^2+(\log_{10}7)^2$?

Hint:

Extended Cauchy-Schwarz Inequality
 
My calculator was faster. (Yes)

-Dan
 
anemone said:
State with reason which of these is greater?

$\log_{10}12$ or $(\log_{10}5)^2+(\log_{10}7)^2$?

My solution:

Note that by using the extended Cauchy-Schwarz inequality, we have

$\dfrac{(\log_{10}5)^2}{1}+\dfrac{(\log_{10}7)^2}{1}\ge \dfrac{(\log_{10}5+\log_{10}7)^2}{1+1}=\dfrac{(\log_{10}35)^2}{2}$

But we know $35^2=1225>10^3$, so $\log_{10}35>\dfrac{3}{2}$ and therefore $\dfrac{(\log_{10}35)^2}{2}>\dfrac{9}{8}$.

If we can prove $\dfrac{9}{8}>\log_{10} 12$, then we can conclude $(\log_{10}5)^2+(\log_{10}7)^2>\log_{10}12$.

Observe that $10^3=1000>864=\dfrac{12^3}{2}$, so we get $10^9>\dfrac{12^9}{8}$, since $\dfrac{12^9}{8}>12^8$, we have proved $10^9>12^8$, i.e. $\dfrac{9}{8}>\log_{10} 12$.

Therefore, $\dfrac{(\log_{10}5)^2}{1}+\dfrac{(\log_{10}7)^2}{1}\ge\dfrac{(\log_{10}35)^2}{2}\ge\dfrac{9}{8}>\log_{10} 12$.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Replies
1
Views
1K
Replies
2
Views
2K
Replies
2
Views
2K
Replies
1
Views
1K
Replies
2
Views
2K
Replies
4
Views
1K
Replies
1
Views
4K
Replies
3
Views
1K
Back
Top