how to find the invserse of
y=3logbase4(3x-2) thanks!
y=3logbase4(3x-2) thanks!
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From the power identity of logarithms, you can rewrite your problem as:
y = logbase4(3x-2)^3
So:
4^y = (3x-2)^3
Solve for x:
(4^y)^(1/3) = 3x - 2
3x = (4^y)^(1/3) + 2
x = ((4^y)^(1/3) + 2)/3
Your inverse is:
y = ((4^x)^(1/3) + 2)/3
y = logbase4(3x-2)^3
So:
4^y = (3x-2)^3
Solve for x:
(4^y)^(1/3) = 3x - 2
3x = (4^y)^(1/3) + 2
x = ((4^y)^(1/3) + 2)/3
Your inverse is:
y = ((4^x)^(1/3) + 2)/3