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

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

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

inverse

f (y) = \frac{1}{3} (2^{3 x + 1}) - 1

\frac{ln ( 3 x + 3 ) - ln ( 2 )}{3 ln ( 2 )}

Solution steps

y = \frac{1}{3} \cdot 2^{3 x + 1} - 1

Replace

x

with

y

x = \frac{1}{3} \cdot 2^{3 y + 1} - 1

Solve for

y, x = \frac{1}{3} 2^{3 y + 1} - 1

\frac{ln ( 3 x + 3 ) - ln ( 2 )}{3 ln ( 2 )}

3 ln (x + 4 - 1), x > 0

f (x) = (x^{3} + 1)^{0.5}

f (x) = \frac{22 x - 14}{1}

f (x) = 10 x^{2} - 2

f (x) = 9.45

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