What is the inverse of $f (x) = 2 ln (x - 1) - i x - x$ $f(x) = 2ln(x-1)-ix-x$ ?

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What is the inverse of $f (x) = 2 ln (x - 1) - i x - x$ $f(x) = 2ln(x-1)-ix-x$ ?

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Answer 1 · 2017-07-05T20:41:27

$f^{- 1} (x) = e^{i x} - x + 1$ $f^(-1)(x)=e^(ix)-x+1$ #

Explanation:

If f" $f (x) = 2 ln (x + 1) - i x - x$ $f(x)=2"ln"(x+1)-ix-x$ , then $f^{- 1} (x)$ $f^(-1)(x)$ is the inverse, or the reflection of $f (x)$ $f(x)$ in the line $y = x$ $y=x$ .

So, let's make $f (x) = 2 ln (x + 1) - i x - x = 0$ $f(x)=2"ln"(x+1)-ix-x=0$ , then $2 ln (x + 1) = i x + x = i (2 x)$ $2"ln"(x+1)=ix+x=i(2x)$ , dividing both sides by 2 gives $f (x) = ln (x + 1) = \frac{i (2 x)}{2} = i x$ $f(x)="ln"(x+1)=(i(2x))/2=ix$ .

As $e^{ln (a)} = a$ $e^("ln"(a))=a$ , $e^{ln (x + 1)} = x + 1 = e^{i x}$ $e^(ln(x+1))=x+1=e^(ix)$ .

Now take away $(x + 1)$ $(x+1)$ from both sides to make it equal 0, $0 = e^{i x} - x + 1$ $0=e^(ix)-x+1$ .

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What is the inverse of $f (x) = 2 ln (x - 1) - i x - x$ $f(x) = 2ln(x-1)-ix-x$ ?

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Explanation:

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What is the inverse of f(x) = 2ln(x-1)-ix-xf(x)=2ln(x−1)−ix−xf(x) = 2ln(x-1)-ix-x?

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What is the inverse of $f (x) = 2 ln (x - 1) - i x - x$ $f(x) = 2ln(x-1)-ix-x$ ?