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

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

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

inverse

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

\frac{5}{( \frac{1}{3} ) ^{x} - 1}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{5}{( \frac{1}{3} ) ^{x} - 1}

f (x) = 3 x

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

y = 3 x + 1, (- 3, 6)

f (x) = \frac{- 5 x - 15}{2 x ^{2} + 5 x - 3}

f (x) = \frac{3}{2} x - 6

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