I was tutoring someone in pre-cal, and I just couldn\'t seem toget the problem.

Question

I was tutoring someone in pre-cal, and I just couldn't seem toget the problem. I got to one point and was stuck fromthere.... It is.... 3 - ln(x) = 2x - 5 from there we have: ln(x) = 8 - 2x or ln(x) + 2x = 8 applying e to both sides we have: xe2x =e8 From here, we're stuck. Ideas or answers? =] We got the answerusing the calculator and graphing, but how can we do it without theuse of a calculator? I was tutoring someone in pre-cal, and I just couldn't seem toget the problem. I got to one point and was stuck fromthere.... It is.... 3 - ln(x) = 2x - 5 from there we have: ln(x) = 8 - 2x or ln(x) + 2x = 8 applying e to both sides we have: xe2x =e8 From here, we're stuck. Ideas or answers? =] We got the answerusing the calculator and graphing, but how can we do it without theuse of a calculator? From here, we're stuck. Ideas or answers? =] We got the answerusing the calculator and graphing, but how can we do it without theuse of a calculator?

Explanation / Answer

We set it up to find the point of intersection of the two graphs y = ln(x) +2x and y = 8 and we found that the x-coordinate of the point of intersection is the answer that is in the back of the book, x = 3.389.

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