Question

log5x +log (2x + 3) = 1 + 2.log(3-x)                x < 3⤵️​

Answer 1

Solution:

Solution:log5x +log (2x + 3) = 1 + 2.log(3-x)

Solution:log5x +log (2x + 3) = 1 + 2.log(3-x)log5x + log(2x + 3) = log10 + log(3-x)2

Solution:log5x +log (2x + 3) = 1 + 2.log(3-x)log5x + log(2x + 3) = log10 + log(3-x)2log(5x.(2x +3)) = log (10.(3-x)2)

Solution:log5x +log (2x + 3) = 1 + 2.log(3-x)log5x + log(2x + 3) = log10 + log(3-x)2log(5x.(2x +3)) = log (10.(3-x)2)5x.(2x +3) = 10.(3-x)2

Solution:log5x +log (2x + 3) = 1 + 2.log(3-x)log5x + log(2x + 3) = log10 + log(3-x)2log(5x.(2x +3)) = log (10.(3-x)2)5x.(2x +3) = 10.(3-x)210x2 +15x = 10.(9-6x + x2)

Solution:log5x +log (2x + 3) = 1 + 2.log(3-x)log5x + log(2x + 3) = log10 + log(3-x)2log(5x.(2x +3)) = log (10.(3-x)2)5x.(2x +3) = 10.(3-x)210x2 +15x = 10.(9-6x + x2)10x2 + 15x = 90-60x +10x2

Solution:log5x +log (2x + 3) = 1 + 2.log(3-x)log5x + log(2x + 3) = log10 + log(3-x)2log(5x.(2x +3)) = log (10.(3-x)2)5x.(2x +3) = 10.(3-x)210x2 +15x = 10.(9-6x + x2)10x2 + 15x = 90-60x +10x275x = 90