What is the inverse of (-1)/2 log_{5}(1/(x+3)) ?

The inverse of (-1)/2 log_{5}(1/(x+3)) is 25^x-3

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inverse of (-1)/2 log_{5}(1/(x+3))

inverse

\frac{- 1}{2} lo g_{5} (\frac{1}{x + 3})

2 5^{x} - 3

Solution steps

y = \frac{- 1}{2} lo g_{5} (\frac{1}{x + 3})

Replace

x

with

y

x = \frac{- 1}{2} lo g_{5} (\frac{1}{y + 3})

Solve for

y, x = \frac{- 1}{2} lo g_{5} (\frac{1}{y + 3})

2 5^{x} - 3

f (x) = \frac{7}{4} x - \frac{2}{5}

f (x) = \frac{6 x + 9}{9 x - 7}

+ f^{9}

f (x) = 2^{2 x^{2} + 2 x} + 3

ln (7 x + 4) - 2

What is the inverse of (-1)/2 log_{5}(1/(x+3)) ?
The inverse of (-1)/2 log_{5}(1/(x+3)) is 25^x-3