What is the inverse of f(x)= 1/3 log_{2}(x)+1 ?

The inverse of f(x)= 1/3 log_{2}(x)+1 is 2^{3x-3}

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inverse of f(x)= 1/3 log_{2}(x)+1

inverse

f (x) = \frac{1}{3} lo g_{2} (x) + 1

2^{3 x - 3}

Solution steps

y = \frac{1}{3} lo g_{2} (x) + 1

Replace

x

with

y

x = \frac{1}{3} lo g_{2} (y) + 1

Solve for

y, x = \frac{1}{3} lo g_{2} (y) + 1

2^{3 x - 3}

x^{2} - x - 2

f (x) = 2 ln (x)

f (x) = \frac{1}{2} x + 1

f (x) = x^{7} + 5

f (x) = (\frac{( 5 x + 31 )}{( 2 x - 13 )})

What is the inverse of f(x)= 1/3 log_{2}(x)+1 ?
The inverse of f(x)= 1/3 log_{2}(x)+1 is 2^{3x-3}