Question

Log(3y+2)-1=log(y-4) solve y​

Answer 1

`\log_(10) (3y+2) - 1 = \log_(10)(y-4)\\\\\implies \log_(10) (3y+2) - \log_(10) 10 = \log_(10) (y-4)\\\\\implies \log_(10) \left( (3y+2)/(10) \right) = \log_(10) (y-4)\\\\\implies (3y+2)/(10) = y-4\\\\\implies 3y+2 = 10y-40\\\\\implies 10y-3y = 40+2\\\\\implies 7y = 42\\\\\implies y = \frac{42}7\\\\\implies y = 6`