Find a formula for the inverse function h^-1(x)

[From: ] [author: ] [Date: 12-11-20] [Hit: ]

(4y+3)/(4y-2) = 10^x = k, 4y+3= 4ky-2k , 3+2k = 4y(k-1) , y = (3+2k)/[4(k-1)]-H(x) is just another way of saying y. The inverse simply means exchange x and y, so where the x was put a y,......

Let h(x)=log(4x+3/4x-2)

I am clueless here. Can you guys please show me the steps? Thank you.

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x = log[(4y+3)/(4y-2)] , (4y+3)/(4y-2) = 10^x = k , 4y+3= 4ky-2k , 3+2k = 4y(k-1) , y = (3+2k)/[4(k-1)]

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H(x) is just another way of saying y. The inverse simply means exchange x and y, so where the x was put a y, and where the y was put an x. So x=log(4y+3/4y-2). Then solve for why. When you are done you can just exchange the y for h(x) again.

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