How do you solve this? Actually it's not homework, I need

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>>8420308
This is a solution. It may not be unique though.

>>8420308
0.7^k = 1.2
k = log 1.2 / log 0.7
(0.7^kx)*(0.7^10-x)
0.7^kx+10-x
0.7^x(k-1)+10 = 1
x(k-1)+10 = 0
x(k-1) = -10
x = -10/(k-1)

>>8420318
>It may not be unique though.

Each value appears only once on an exponential curve, unless you include complex numbers.

>>8420318
Hey thanks a lot for the help. I understand how you got 1.2^x = 0.7^(-10+x)

But what if it wasn't that convenient? What if this equation = 2 instead of 1

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>>8420343
General case in pic related for any real numbers a,b and c and any real functions f and g

You seem to be doing real math at your job but you don't seem to know that much. May I ask what you do and what your education is in?

>>8420343
Then you'd just have 1.2^x=2*.7^(-10+x)
When you take log base 1.2 of both sides, you get
x=log(2*.7^(-10+x))
this log can then be split up into two logs
x=log(2)+log(.7^(-10+x))
Same question as the other anon, what is your job? I thought this had something to do with probability at first but then I realized 1.2 can't be a probability.

>>8420350
i don't work a job. this is about trading derivatives and stocks. i havent studied math since highschool, so whenever i find a math problem in front of me, i have to look up everything. It makes me kind of wish I took math instead of some nursing bullshit

in this particular situation, i am trying to figure out how many trades i would have to succeed and fail to break even given that i am trading with my entire account each time and my average gain is 1.2x my account, average loss is 0.7x. i set the number of trials to 10 for the sake of simplicity and understanding how to solve this. now you should understand how i came up with this equation.

the next step is for me to use cumulative binomial distribution to figure out what is my chance of breaking even or doing better. in that case, i now know the value of x, along with n and p, so i can just punch that in the formula and it will tell me what is p(X>=x).

This is redundant with a high probability of success and gains. But for strategy with high probability of failure and loss, I have to do this to somehow figure out how much I should be betting each time to avoid my account going below a certain value. I haven't got to that step yet

>>8420361
It is about probability.

>>8420371
I see. I guess it is true that doing math to then get into finance is a good idea.

That said, playing with your own money doesn't sound like fun but given that you studied nursing I can understand why you'd do it.

>>8420378
Okay I see that now. I think when using the binomial distribution, you should make a program on excel to check for multiple different values of n. That would give you a better idea for the shape of the curve. Also, instead of using average losses and average gains, a more accurate way would be to find the probability of a given loss or gain, then using the formula in the OP then multiplying the two. This would result in an infinite sum, but sometimes it's good to add up the most important terms in order to get a better idea of what's actually going on. I'm only trying to give you different techniques and advice since you're already playing around with the numbers, I just want to make sure you're playing in the most effective way. Good luck.

>>8420318
1.2 = .7 ^ (log.7(1.2))
1.2^x = .7 ^ [x * log.7(1.2)]
(1.2^x)(0.7^(10-x)) = .7 ^ [x*log.7(1.2) + 10 - x]
0=log.7(1)=x*(log.7(1.2) - 1) + 10
-10 / (log.7(1.2) - 1) = x

Wew lad, your process seems a little fucky :(

[math]
x = \log_{\frac{1.2}{0.7}}\left(0.7^{10} \right) \approx 6.617\dots
[/math]

>>8420441
Correction [math] 0.7^{-10} [/math].

See https://www.wolframalpha.com/input/?i=x+%3D+Log%5B1.2%2F0.7,+0.7%5E(-10)%5D;+1.2%5Ex+*+0.7%5E(10-x).

In the form of
A = b^x * c^(n-x)

x = Logb(c) - Logb(A) / (Logb(c) - 1)

Any way to simplify?

>>8420585
I forgot n
x =n* Logb(c) - Logb(A) / (Logb(c) - 1)

>>8420441
How did that solution come out?

>>8420371
>trading derivatives and stocks
kys no really kys. Gold price is already being manipulated. Buy and hide. Off to >>/biz/