SOLUTION: expess 1/5 ln(x + 2)^5 + 1/2 [ln x − ln(x^2 + 3x + 2)^2] as a single logarithm
i got 1/10 log (x(x+2)^2)/(x^2+3x+2)^2) wrong
and then i tried ln((x+2)x^1/2)/ (x^2+3x+2)
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-> SOLUTION: expess 1/5 ln(x + 2)^5 + 1/2 [ln x − ln(x^2 + 3x + 2)^2] as a single logarithm
i got 1/10 log (x(x+2)^2)/(x^2+3x+2)^2) wrong
and then i tried ln((x+2)x^1/2)/ (x^2+3x+2)
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Question 804240: expess 1/5 ln(x + 2)^5 + 1/2 [ln x − ln(x^2 + 3x + 2)^2] as a single logarithm
i got 1/10 log (x(x+2)^2)/(x^2+3x+2)^2) wrong
and then i tried ln((x+2)x^1/2)/ (x^2+3x+2) wrong also Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! expess 1/5 ln(x + 2)^5 + 1/2 [ln x − ln(x^2 + 3x + 2)^2] as a single logarithm
i got 1/10 log (x(x+2)^2)/(x^2+3x+2)^2) wrong
and then i tried ln((x+2)x^1/2)/ (x^2+3x+2) wrong also
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expess 1/5 ln(x + 2)^5 + 1/2 [ln x − ln(x^2 + 3x + 2)^2] as a single logarithm
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1/5 ln(x + 2)^5 + 1/2 [ln x - ln(x^2 + 3x + 2)^2]
= ln(x+2) + (1/2)ln(x) - ln(x^2 + 3x + 2)
= ln(sqrt(x)*(x+2)/(x^2 + 3x + 2))
= ln(sqrt(x)*(x+2)/((x + 2)*(x+1))
=
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and then i tried ln((x+2)x^1/2)/ (x^2+3x+2) wrong also
that's not wrong, but you didn't cancel a factor.
