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

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

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

inverse

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 )}

2 x^{2} - 4

y = \frac{4 x + 4}{x - 4}

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

f (x) = 7 x + 9 x + 2

tan (- 3)

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