What is the inverse of $y = {log}_{4} (x - 3) + 2 x ?$ $y=log_4( x-3) +2x?$ ?

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Answer 1 · 2018-02-26T09:09:35

$x = \frac{1}{2} (6 + W (2^{2} y - 11))$ $x = 1/2(6+W(2^2y-11))$

We can solve this problem using the so called Lambert function $W (\cdot)$ $W(cdot)$

$y = \frac{ln | x - 3 |}{ln 4} + 2 x \Rightarrow y ln 4 = ln | x - 3 | + 2 x ln 4$ $y = lnabs(x-3)/ln4 + 2x rArr y ln4=lnabs(x-3)+2x ln4$

Now making $z = x - 3$ $z = x-3$

$e^{y ln 4} = z e^{2 (z + 3) ln 4} = z e^{2 z} e^{6 ln 4}$ $e^(y ln4)=z e^(2(z+3)ln4) = ze^(2z)e^(6 ln4)$ or

$e^{(y - 6) ln 4} = z e^{2 z}$ $e^((y-6)ln4)=z e^(2z)$ or

$2 e^{(y - 6) ln 4} = 2 z e^{2 z}$ $2 e^((y-6)ln4) = 2z e^(2z)$

Now using the equivalence

$Y = X e^{X} \Rightarrow X = W (Y)$ $Y = X e^X rArr X = W(Y)$

$2 z = W (2 e^{(y - 6) ln 4}) \Rightarrow z = \frac{1}{2} W (2 e^{(y - 6) ln 4})$ $2z=W(2 e^((y-6)ln4)) rArr z = 1/2 W(2 e^((y-6)ln4))$

and finally

$x = \frac{1}{2} W (2 e^{(y - 6) ln 4}) + 3$ $x = 1/2 W(2 e^((y-6)ln4))+3$ that can be simplified to

$x = \frac{1}{2} (6 + W (2^{2 y - 11}))$ $x = 1/2(6+W(2^(2y-11)))$

What is the inverse of y=log4(x−3)+2x?y=log_4( x-3) +2x? ?