What is the inverse of y=(1/2)^{4-3x}-7 ?

The inverse of y=(1/2)^{4-3x}-7 is (ln(16(x+7)))/(3ln(2))

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inverse of y=(1/2)^{4-3x}-7

inverse

y = (\frac{1}{2})^{4 - 3 x} - 7

\frac{ln ( 16 ( x + 7 ) )}{3 ln ( 2 )}

Solution steps

y = (\frac{1}{2})^{4 - 3 x} - 7

Replace

x

with

y

x = (\frac{1}{2})^{4 - 3 y} - 7

Solve for

y, x = (\frac{1}{2})^{4 - 3 y} - 7

\frac{ln ( 16 ( x + 7 ) )}{3 ln ( 2 )}

f (t) = \frac{4 - t ^{2}}{2 - t}

- \frac{3}{2} sin (2 x - \frac{3 π}{4}) + \frac{7}{3}

f (x) = (0, - 5) (5, 0)

f (x) = 2 (2)^{4} - 16 (2)^{3} + 8

(1, 3), (2, 5)

What is the inverse of y=(1/2)^{4-3x}-7 ?
The inverse of y=(1/2)^{4-3x}-7 is (ln(16(x+7)))/(3ln(2))