Obel1x wrote: Prove that the following inequality holds: \[ \left( \log_{24}48 \right)^2+ \left( \log_{12}54 \right)^2>4\] bigger then $1.2^2+1.6^2=2^2$

SCP wrote: Obel1x wrote: Prove that the following inequality holds: \[ \left( \log_{24}48 \right)^2+ \left( \log_{12}54 \right)^2>4\] bigger then $1.2^2+1.6^2=2^2$ How do you get that?

myro111 wrote: SCP wrote: Obel1x wrote: Prove that the following inequality holds: \[ \left( \log_{24}48 \right)^2+ \left( \log_{12}54 \right)^2>4\] bigger then $1.2^2+1.6^2=2^2$ How do you get that? calculating: 24^{1.2}<24*32^{0.2}=48$ and similar.

actually $LHS=4.060718870...>4$.

Note that $2^{5}>24$ thus $\log_{24}2>0.2$ and \[ 4.5^{5}=(4+0.5)^{5}>4^{5}+5*4^{4}*0.5+10*4^{3}*0,5^{2}=1024+2.5*256+2.5*64=1024+640+160=1824>1728=12^{3}\] so $\log_{12}4.5>\frac{3}{5}=0.6$ Now $(\log_{24}48)^{2}+(\log_{12}54)^{2} = (\log_{24}(24*2))^{2}+(\log_{12}(12*4.5))^{2}$ $=(\log_{24}24+\log_{24}2)^{2}+(\log_{12}12+\log_{12}4.5)^{2}=(1+\log_{24}2)^{2}+(1+\log_{12}4.5)^{2}$ $>1.2^{2}+1.6^{2}=1.44+2.56=4$ as required

Kosovo 2011 Obel1x wrote: Prove that the following inequality holds: \[ \left( \log_{24}48 \right)^2+ \left( \log_{12}54 \right)^2>4\] $2^5>24\implies48>24^{\frac{6}{5}}$ and $3^7=2187>2048=2^{11}\implies54>12^{\frac{8}{5}}$, $(\log_{24}48 )^2 +(\log_{12}54)^2 >\left(\frac{6}{5}\right)^2+\left(\frac{8}{5}\right)^2=4$.