How to Check Your Answers for Question 1 and Question 2 - Quick Tips
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To check answers for Question 1, it's important to follow each step with a new equation, starting from xlog x = 100x and applying the logarithmic property log(ab) = log(a) + log(b). The addition of log x in the equation is necessary for proper simplification. For Question 2, the answer should be rounded to 34.7 years, and it's crucial to use "Ln" for natural logarithm instead of "In." Posting the problem and work directly can facilitate quicker assistance.
Q1: Your work is incorrect.
You're starting with an equation, so each step should be a new equation.
xlog x = 100x
log(xlog x) = log(100x)
(log x)(log x) = log 100 + log x
Continue from there. You should end up with a value of x that is correct to three significant digits.
For Q2, your answer is close, but should be rounded to 34.7 years, to the nearest tenth of a year.
Also, there is no In operation. This is Ln, and stands for natural Logarithm, which in Latin is logarithmus naturalis.
BTW, you might get quicker help if you posted the problem and your work rather than posting a scanned copy of your work. Speaking for myself, it's much easier when I can write my response while looking at the work, rather than having to have two windows open and jump back and forth between them.
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached.
The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter.
From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers:
Show that, for ##|r|<1,##
$$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$
My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above:
$$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$
The sum of this series is just:
$$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$
I'm having some trouble trying to figure out what to...