[tex]3)\log_{49}7=\log_{7^2}7=\dfrac{1}{2} \log_7 7=\dfrac{1}{2} \times1=\dfrac{1}{2} \\\\4)\log40=\log(10\times4)=\log_{10}10+\log_{10}2^2=1+2\log_{10}2\\\log_{10}2<0.5 =>2\log_{10}2<1\\=>\log40\approx1.602[/tex]
