What is the inverse of f(x)=log_{6}(8x)-log_{6}(3) ?

The inverse of f(x)=log_{6}(8x)-log_{6}(3) is 2^{x-3}*3^{1+x}

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inverse of f(x)=log_{6}(8x)-log_{6}(3)

inverse

f (x) = lo g_{6} (8 x) - lo g_{6} (3)

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

Solution steps

y = lo g_{6} (8 x) - lo g_{6} (3)

Replace

x

with

y

x = lo g_{6} (8 y) - lo g_{6} (3)

Solve for

y, x = lo g_{6} (8 y) - lo g_{6} (3)

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

y = 4 x + 3

f (x) = \frac{2 x + 1}{3 x ^{2} - 39 x + 108}

\frac{3}{x ^{2} - 16}

(6, 2), (- 6, - 2)

\frac{8}{7 + x}

What is the inverse of f(x)=log_{6}(8x)-log_{6}(3) ?
The inverse of f(x)=log_{6}(8x)-log_{6}(3) is 2^{x-3}*3^{1+x}