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Solve ln (x+1)=ln x+1 for x
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This might make it look easier: \(ln(x+1) = ln(x) + ln(e) \)
move logs to one side combine logs \[\ln (x+1) -\ln x = 1\] \[\ln \frac{x+1}{x} = 1\] take "e" raised to both sides or change to exponential form \[\frac{x+1}{x} = e\] \[x+1 = ex\] \[1 = ex -x\] \[x = \frac{1}{e-1}\]
@tkhunny , that would save a couple of steps, i didn't think of that
Thank you both
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