Question

If log2 + log(x+3) - log(3x-5) = log3. The value of x= ?
A -3

Answer 1

$\rm{\red{Answer}}$

$\boxed{\rm{\pink{3}}}$

$\mathbb{GIVEN\:QUESTION\:IS}$

$\rm{\log_{e}2+\log_{e}\left(3+x\right)-\log_{e}\left(3x-5\right)=\log_{e}\left(3\right)}$

$\underline{\mathbb{EXPLANATION}}$

$\implies{\rm{\log_{e}\left(2x+6\right)-\log_{e}\left(3x-5\right)=\log_{e}\left(3\right)}}$

$\boxed{\rm{\pink{Becoz\:\:\log_{e}m+\log_{e}n=\log_{e}\left(mn\right)}}}$

$\implies{\rm{\log_{e}\left(2x+6\right)-\log_{e}\left(3x-5\right)-\log_{e}\left(3\right)=0}}$

$\implies{\rm{\log_{e}\left(2x+6\right)-\big(\log_{e}\left(3x-5\right)+\log_{e}\left(3\right)\big)=0}}$

$\implies{\rm{\log{e}\left(2x+6\right)-\log_{e}\left(9x-15\right)=0}}$

$\implies{\rm{\log_{e}\left(\frac{2x+6}{9x-15}\right)=0}}$

$\boxed{\rm{\pink{Becoz\:\:\log_{e}m-\log_{e}n=\log_{e}\left(\frac{m}{n}\right)}}}$

$\implies{\rm{\left(\frac{2x+6}{9x-15}\right)=e^0}}$

$\boxed{\rm{\pink{Becoz\:\:IF\:log_{e}x=y\:Then\:\:x=e^y}}}$

$\implies{\rm{\left(\frac{2x+6}{9x-15}\right)=1}}$

$\implies{\rm{2x+6=9x-15}}$

$\implies{\rm{2x-9x=-15-6}}$

$\implies{\rm{-7x=-21}}$

$\implies{\rm{x=\frac{21}{7}}}$

$\implies{\rm{x=\frac{7\times3}{7}}}$

$\implies{\rm{x=3}}$