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Natural Logarithm and Integral Properties

I was asked to prove that ln(xy) = ln x + ln y using the integral definition. While I'm not asking for any answers on the proof, I was wondering how to interpret and set-up this proof using the "integral definition" (As I am unsure what that means.)

EDIT

And to prove that ln(x/y) = ln x - ln y

Is it right to say this?

$\ln(\frac{x}{y})=\int_1^{\frac{x}{y}} \frac{dt}{t}=\int_1^x \frac{dt}{t}-\int_x^{\frac{x}{y}}\frac{dt}{t}.$

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    [This wikipedia page](http://en.wikipedia.org/wiki/Natural_logarithm#De$f$initions) might help you...2012-09-16

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By definition, $\ln w=\int_1^w \frac{dt}{t}.$ Thus $\ln(xy)=\int_1^{xy} \frac{dt}{t}=\int_1^x \frac{dt}{t}+\int_x^{xy}\frac{dt}{t}.$ Now make an appropriate change of variable to conclude that the last integral on the right is equal to $\ln y$.

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    @Johannes: No point in starting over. Note that $(x/y)(y)=x$. so by the above result $\ln(x)=\ln(x/y)+\ln(y)$. Basically finished!2012-09-16