What is the inverse of 9^x+2*3^x+3,x<= 0 ?

The inverse of 9^x+2*3^x+3,x<= 0 is (ln(-1+sqrt(x-2)))/(ln(3)),(ln(-sqrt(x-2)-1))/(ln(3))

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inverse of 9^x+2*3^x+3,x<= 0

inverse

9^{x} + 2 \cdot 3^{x} + 3, x \leq 0

\frac{ln ( - 1 + x - 2 )}{ln ( 3 )}, \frac{ln ( - x - 2 - 1 )}{ln ( 3 )}

Solution steps

y = 9^{x} + 2 \cdot 3^{x} + 3

Replace

x

with

y

x = 9^{y} + 2 \cdot 3^{y} + 3

Solve for

y, x = 9^{y} + 2 \cdot 3^{y} + 3

\frac{ln ( - 1 + x - 2 )}{ln ( 3 )}, \frac{ln ( - x - 2 - 1 )}{ln ( 3 )}

\frac{8 x}{3 x - 5}

f (x) = (x^{8} - 9)

f (x) = 4 x - 5 - 10

\frac{( - 4 x - 4 )}{7 x + 8}

f (x) = \frac{( 12 - 4 x )}{( 12 - x )}

What is the inverse of 9^x+2*3^x+3,x<= 0 ?
The inverse of 9^x+2*3^x+3,x<= 0 is (ln(-1+sqrt(x-2)))/(ln(3)),(ln(-sqrt(x-2)-1))/(ln(3))