What is log_{4}(2x+1)-log_{4}(3y+4) ?

The solution to log_{4}(2x+1)-log_{4}(3y+4) is log_{4}((2x+1)/(3y+4))

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simplify log_{4}(2x+1)-log_{4}(3y+4)

simplify

lo g_{4} (2 x + 1) - lo g_{4} (3 y + 4)

lo g_{4} (\frac{2 x + 1}{3 y + 4})

Solution steps

lo g_{4} (2 x + 1) - lo g_{4} (3 y + 4)

Apply log rule:

lo g_{a} (x) - lo g_{a} (y) = lo g_{a} (\frac{x}{y})

lo g_{4} (2 x + 1) - lo g_{4} (3 y + 4) = lo g_{4} (\frac{2 x + 1}{3 y + 4})

= lo g_{4} (\frac{2 x + 1}{3 y + 4})

ln (\frac{1}{2 e})

lo g_{7} (2 x - 5) + lo g_{7} (5)

lo g_{10} (\frac{1}{ab})

- 4 ln (2 x)

- \frac{5 ln ( 50 )}{ln ( 2 )}

What is log_{4}(2x+1)-log_{4}(3y+4) ?
The solution to log_{4}(2x+1)-log_{4}(3y+4) is log_{4}((2x+1)/(3y+4))