Probably not a hard question for all you maths bofffins out there but...
If 1/x = x^-1 , 2/x = x^what? I need an answer quickly if possible. Any help or links are appreciated.
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Probably not a hard question for all you maths bofffins out there but...
If 1/x = x^-1 , 2/x = x^what? I need an answer quickly if possible. Any help or links are appreciated.
I have discovered that:
if
y=n/x then n = (logx/logy)
how can this be adapted. I don't know what y is. Is there such a function as an inverse log?
OK. I've finally figured it out. I think.
If y=N/x
Then
y= x^(logx/log(N/x))
Am I right?
Indicies.
1/x = x^-1
2/x = 2x^-1
1/x^2 = x^-2
2/x^2 = 2x^-2
...and so on.
Logs
log(y^x) = x*log(y)
This is one of a few rules of Logs.
Logs can be used in the problems where you need to find unknown powers. ie.
Find x. when 8^x = 64
log(8^x) = log(64)
x*log(8) = log(64)
x = log(64)/log(8)
x = 2
Hope this clears a few things up. Post back if you need more information. :)
N / x = N * x^ -1
In general N/ (x^K) = N * x^ -K
Inverse log is an exponential
If y = e^x then x = ln(y)
If y = ln(x) then x = e^y
If y = 10^x then x = Log(y)
If y = Log(x) then x = 10^y
Your notions about logarithms are suspect.
If y = N / x then Log(y) = Log(N) - Log(x)
If y = N * x then Log(y) = Log(N) + Log(x)
If y = x^N then Log(Y) = N * Log(x)
Thanks for that. I was trying to be too complicated. I'm trying to fit this formula to find the gradient of a tangent of a curve of the formula:
y = x^n
because I knew that a reciprocal curve (y=N/x) was y=x^n then I thought it would fit.
My formula for the y=x^n curve was:
g=nx^(n-1)
for the reciprocal I tried the same formula but with
n = (log(N/x)/logx).
After some quick sketches I found the formula did not work.
Anyway, thanks. I owe you one.