MHB Logarithm + Nepper's number Exercise

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To solve the equation 2e^(-x) = 3e^(0.1x), the first step is to manipulate the equation to isolate the exponential term. By multiplying through by e^x, the equation simplifies to 2 = 3e^(1.1x). Dividing both sides by 3 leads to (2/3) = e^(1.1x), which can be converted to logarithmic form as 1.1x = ln(2/3). Finally, solving for x gives x = ln(2/3) / 1.1. The discussion also touches on using LaTeX for formatting mathematical expressions.
Velo
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" Solve the equation:

2e^(-x) = 3e^(0.1x) "

I've been fiddling around with this and I have no idea what I'm supposed to do. I know the final answer should be something like:
x = [(2/3)log e] / 1,1

The only step I've managed to do was:
(2/3) x e^(-x) = e^(0.1x)

But after that, I don't know how to get the x out of the exponential.. Help would be appreciated :')
 
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We are given to solve:

$$2e^{-x}=3e^{0.1x}$$

My first step would be to multiply though by $e^x\ne0$ to get:

$$2=3e^{1.1x}$$

Now divide though by 3:

$$\frac{2}{3}=e^{1.1x}$$

Next, what do you get when converting from exponential to logarithmic form?
 
Hm...

1.1x = log e ^ (2/3) <=>
x = [log e ^ (2/3)]/1.1 <=>
x = [(2/3) log e]/1.1

Which was the solution :o Is that right? Also, sorry, I'm having trouble using the latex thingy ><
 
Velo said:
Hm...

1.1x = log e ^ (2/3) <=>
x = [log e ^ (2/3)]/1.1 <=>
x = [(2/3) log e]/1.1

Which was the solution :o Is that right? Also, sorry, I'm having trouble using the latex thingy ><

Let's go back to:

$$\frac{2}{3}=e^{1.1x}$$

Now recall that:

$$a=b^c\implies c=\log_b(a)$$

And so we may write:

$$1.1x=\ln\left(\frac{2}{3}\right)$$

And then on dividing though by 1.1, we get:

$$x=\frac{\ln\left(\dfrac{2}{3}\right)}{1.1}$$

To use $\LaTeX$, you need to enclose your code with tags. The simplest way is to click the $\Sigma$ button on our toolbar, and then the cursor will be located in between the resulting $$$$ tags, and you can add your code there. You will find most symbols/commands you need to the right of the editor, in the "Quick $\LaTeX$" tool, and you can get quick previews of your code in our "$\LaTeX$ Live Preview" tool.
 
Ohhh, I get it now.. I read the book wrong too :') The solution had $$ \log_e(\frac{2}{3}) $$ and not $$\log(\frac{2}{3})$$.. I spent so much time wondering where that log had come from :') I tried redoing that exercise and the next in my notebook and I'm doing alright now x3 Thank you very much! :D
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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