What is the inverse of log_{10}(2/(x+1)) ?

The inverse of log_{10}(2/(x+1)) is (2^{1-x})/(5^x)-1

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inverse of log_{10}(2/(x+1))

inverse

lo g_{10} (\frac{2}{x + 1})

\frac{2 ^{1 - x}}{5 ^{x}} - 1

Solution steps

y = lo g_{10} (\frac{2}{x + 1})

Replace

x

with

y

x = lo g_{10} (\frac{2}{y + 1})

Solve for

y, x = lo g_{10} (\frac{2}{y + 1})

\frac{2 ^{1 - x}}{5 ^{x}} - 1

f (x) = \frac{( 4 x - 2 )}{3 x + 1}

f (x) = x^{2} - 8 x + 17

f (x) = x^{2} - 8 x + 18

6 x + 7

\frac{1}{10 - 5}

What is the inverse of log_{10}(2/(x+1)) ?
The inverse of log_{10}(2/(x+1)) is (2^{1-x})/(5^x)-1