$\begingroup$ To find inverse of y=f(x), we first re-write the equation as x=g(y), and then replace x with y, to get the inverse as y=g(x).(this replacement of y with x is done because the function and its inverse are symmetric about the line y=x) $\endgroup$ –

2018年10月1日 · The question demands an answer to when ln(x+y) can be considered to equal ln(x)+ ln(y) and we are asked to use contour to prove this but the concept of contour does not entirely make sense to me. Anyways the answer is for y=x/(x-1) …

As you asked for a hint ( good job) If you see the definition of $\ln(x)$ you can notice that it's the inverse function of $ e^x$ then follow the law of exponents and take the inverse function on both sides.

In my case the result of y is always a real positive number (y>=1). Therefore, I would like to avoid any solutions that contain any imaginary part. problem-solving

2011年11月10日 · If X is an exponential random variable with parameter \lambda =1, find the probability density function of the random variable Y=lnX. HINT: Find the cummulative distribution function. I don't understand what the hint saying, find the cdf of what? when \lambda =1, we have f(x)=e^{-x} , Y=lnX implies X=e^y , if I sub e^y to f(x), it will be very ...

2016年7月6日 · While attempting the problem below, I got the cross product of the function's derivative and second derivative to equal 0. If this happens, what does this mean for the curvature formula? My work: ...

2019年2月1日 · Find the volume of the region bounded by the graphs of y=lnx, y=0, and x=4 rotated about x=5. Ask Question ...

2019年2月1日 · $\begingroup$ Hint: Take the natural logarithm on both sides of the equation which gives you $\ln (y)=\ln(x) \ln(\ln (x))$. Now all you have to do is to implicitly differentiate and use the product rule and the chain rule. $\endgroup$

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2017年10月4日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.