What is the inverse of 2x=ln((e^{3y}-4)/3) ?

The inverse of 2x=ln((e^{3y}-4)/3) is (ln((e^{3x}-4)/3))/(2)

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inverse of 2x=ln((e^{3y}-4)/3)

inverse

2 x = ln (\frac{e ^{3 y} - 4}{3})

\frac{ln ( \frac{e ^{3 x} - 4}{3} )}{2}

Solution steps

y = \frac{ln ( e ^{2 x} \cdot 3 + 4 )}{3}

Replace

x

with

y

x = \frac{ln ( e ^{2 y} \cdot 3 + 4 )}{3}

Solve for

y, x = \frac{ln ( e ^{2 y} 3 + 4 )}{3}

\frac{ln ( \frac{e ^{3 x} - 4}{3} )}{2}

\frac{e ^{x} - 5}{4 - e ^{x}}

f (x) = x - 20

f (x) = x^{2} - 6 x + 2

f (x) = 1 0^{x - 2} + 10

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

What is the inverse of 2x=ln((e^{3y}-4)/3) ?
The inverse of 2x=ln((e^{3y}-4)/3) is (ln((e^{3x}-4)/3))/(2)