For ln√[x³/(x+1)], I rewrote it as ln[x³/(x+1)]^½. From there, do I use the difference property of logarithm (ln(x/y)=lnx-lny)?
And then for 2ln3-½ln(x²+1), I need to simplify it to a single quantity, but I'm unsure of where to start
And then for 2ln3-½ln(x²+1), I need to simplify it to a single quantity, but I'm unsure of where to start
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1)
ln√[x³/(x+1)]
= 1/2 ln (x^3 / (x + 1)
= (1/2) ln x^3 - 1/2 ln (x + 1)
= (3/2) ln x - (1/2) ln (x + 1)
2)
2 ln 3 - ½ ln(x²+1)
= ln(3)^2 - ln√(x^2 + 1)
= ln[ 9 /√(x^2 + 1)]
ln√[x³/(x+1)]
= 1/2 ln (x^3 / (x + 1)
= (1/2) ln x^3 - 1/2 ln (x + 1)
= (3/2) ln x - (1/2) ln (x + 1)
2)
2 ln 3 - ½ ln(x²+1)
= ln(3)^2 - ln√(x^2 + 1)
= ln[ 9 /√(x^2 + 1)]
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ln[x³/(x+1)]^½ = (1/2)(3 lnx - ln(x+1)) , you can't simplify to the single ln