What is the inverse of log_{4}(7x-5)-log_{4}(3x-8) ?

The inverse of log_{4}(7x-5)-log_{4}(3x-8) is (-5+2^{3+2x})/(3*4^x-7)

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inverse of log_{4}(7x-5)-log_{4}(3x-8)

inverse

lo g_{4} (7 x - 5) - lo g_{4} (3 x - 8)

\frac{- 5 + 2 ^{3 + 2 x}}{3 \cdot 4 ^{x} - 7}

Solution steps

y = lo g_{4} (7 x - 5) - lo g_{4} (3 x - 8)

Replace

x

with

y

x = lo g_{4} (7 y - 5) - lo g_{4} (3 y - 8)

Solve for

y, x = lo g_{4} (7 y - 5) - lo g_{4} (3 y - 8)

\frac{- 5 + 2 ^{3 + 2 x}}{3 \cdot 4 ^{x} - 7}

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

2 (n - 2)^{3}

f (x) = - \frac{4}{3} x + \frac{8}{3}

f (x) = 6 x^{3} + 4

f (x) = 6 x^{3} + 7

What is the inverse of log_{4}(7x-5)-log_{4}(3x-8) ?
The inverse of log_{4}(7x-5)-log_{4}(3x-8) is (-5+2^{3+2x})/(3*4^x-7)