1

If $abc= 2^6$, $a, b, c \ge 0$, $\log_2 (a)\log_2 (bc)+\log_2 (b)\log_2 (c)= 10$, find $\sqrt{((\log_2 (a))^2 + (\log_2 (b))^2 + (\log_2 (c))^2}$

David HM
  • 11
  • 2
  • Let $A=\log_2 (a)$ and similarly for $B,C$. Expand the logs and replace the variables. You have only two equations in three unknowns, so should not expect a unique solution. I believe there is only one solution in the integers, but if that is what you are looking for, you should say that. In that case a simple search will find it. – Ross Millikan Jan 15 '20 at 23:01
  • $\log_2(a^{\log_2(b)}+a^{\log_2(c)})+log_2(b^c)=10$ –  Jan 15 '20 at 23:18

1 Answers1

2

Using the substitution suggested in the comments gives the equations \begin{cases} A+B+C=6\\ AB+AC+BC=10 \end{cases} We can then square the first equation to give \begin{align} (A+B+C)^2 &=A^2+B^2+C^2+2(AB+AC+BC)\\ &=A^2+B^2+C^2+2\cdot10\\ &=A^2+B^2+C^2+20\\ &=6^2\\ &=36\\ \end{align} Thus we get $$A^2+B^2+C^2=36-20=16$$ Note that you want the quantity $$\sqrt{A^2+B^2+C^2}=\sqrt{16}=4$$

Peter Foreman
  • 19,947
  • Thanks. The answer key is = 4. Now let me work out with your clue. Hopefully I can find the steps – David HM Jan 15 '20 at 23:26
  • Thank you so much for your kindness, especially @Peter Foreman for the squaring the first equation. I've worked out the complete steps – David HM Jan 15 '20 at 23:40