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spartas
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find the inverse function of
f(x)=ln(x-1), x>1
f(x)=ln(x-1), x>1
MHB Find Inverse of f(x) = ln(x-1)
- Thread starter spartas
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To find the inverse of the function f(x) = ln(x-1) for x > 1, set y = f(x). This leads to the equation y = ln(x-1), which can be rewritten as e^y = x - 1. Solving for x gives x = e^y + 1. Therefore, the inverse function is f^(-1)(y) = e^y + 1.
Obviously, there is something elementary I am missing here.
To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix
in the real plane; taking the transpose we get
which then corresponds to a-bi...
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