3ln(y) = (1/2)ln(2x+1) - (1/3)ln(x+4) + ln(C)?

Contract, expressing your answer as a single logarithm. Note that C is a constant.





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3ln(y) = (1/2)ln(2x+1) - (1/3)ln(x+4) + ln(C)?
3 ln y = 1/2 ln (2x+1) - 1/3 ln (x+4) + ln C


ln y = 1/6 ln (2x+1) - 1/9 ln (x+4) + 1/3 ln (c)


ln y = ln (2x+1)^1/6 - ln (x+4)^1/9 + ln C^1/3


ln y = ln {[(2x+1)^1/6]/[(x+4)^1/9]*C^1/3}


y = [(2x+1)^1/6]/[(x+4)^1/9]*C^1/3
Reply:3 ln y = 1/2 ln (2x+1) - 1/3 ln (x+4) + ln C


ln y = 1/6 ln (2x+1) - 1/9 ln (x+4) + 1/3 ln (c)


ln y = ln (2x+1)^1/6 - ln (x+4)^1/9 + ln C^1/3


ln y = ln {[(2x+1)^1/6]/[(x+4)^1/9]*C^1/3}


y = [(2x+1)^1/6]/[(x+4)^1/9]*C^1/3

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