What is the inverse of log_{9}(x)+log_{9}(12) ?

The inverse of log_{9}(x)+log_{9}(12) is 1/4*3^{-1+2x}

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inverse of log_{9}(x)+log_{9}(12)

inverse

lo g_{9} (x) + lo g_{9} (12)

\frac{1}{4} \cdot 3^{- 1 + 2 x}

Solution steps

y = lo g_{9} (x) + lo g_{9} (12)

Replace

x

with

y

x = lo g_{9} (y) + lo g_{9} (12)

Solve for

y, x = lo g_{9} (y) + lo g_{9} (12)

\frac{1}{4} \cdot 3^{- 1 + 2 x}

y = - 15 x^{2} + 350 x - 2000

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

y = ln (x - 6)

0 = 4 x^{2} - 28 x + 16 y - 31

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

What is the inverse of log_{9}(x)+log_{9}(12) ?
The inverse of log_{9}(x)+log_{9}(12) is 1/4*3^{-1+2x}