What is the inverse of 1/logX?

Question

Also prove that 1/logx has an inverse!

Answer 1

I think I did it. If y=1/log(x), then a few manipulations gave me

x=10^(1/y).

I'm assuming by log you mean log to the base 10. I suppose mathematicians will start fussing about what happens at y=0, but a physicist like me doesn't worry that much about infinities ...

Answer 2

Would the answer not be so simple as logX, in the same way as the inverse of 1/3 is equal to 3?

Answer 3

A. What is the inverse of 1/logX?

y=1/logX; I will assume this is log to base e
logX=1/y, for y not equal to zero
e^(logX) = e^(1/y)
X = e^(1/y).

B. Also prove that 1/logx has an inverse.

y= 1/logX = 1/log(e^(1/y)) = 1/(1/y) = y.

If the base is some number b other than e, just substitute b for e above.

Answer 4

A. what's the inverse of a million/logX? y=a million/logX; i will anticipate this is log to base e logX=a million/y, for y no longer equivalent to 0 e^(logX) = e^(a million/y) X = e^(a million/y). B. additionally instruct that a million/logx has an inverse. y= a million/logX = a million/log(e^(a million/y)) = a million/(a million/y) = y. If the backside is a few variety b different than e, purely replace b for e above.

Answer 5

log a2 to the base a1 is 1/ log a1 to the base a2.

Also we now that the logarithmic function y = log x (base a) is the inverse of y = a ^x.

Therefore the function y = log x( base a1) (= 1/ log a1 to the base x) is the inverse of y = a1 ^ x.

Answer 6

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