What is the inverse of f(x)=log_{10}(-2x^2+8x-a) ?

The inverse of f(x)=log_{10}(-2x^2+8x-a) is -(-8+sqrt(64+8(-a-10^x)))/4 ,-(-8-sqrt(64+8(-a-10^x)))/4

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inverse of f(x)=log_{10}(-2x^2+8x-a)

inverse

f (x) = lo g_{10} (- 2 x^{2} + 8 x - a)

- \frac{- 8 + 64 + 8 ( - a - 1 0 ^{x} )}{4}, - \frac{- 8 - 64 + 8 ( - a - 1 0 ^{x} )}{4}

Solution steps

u = lo g_{10} (- 2 x^{2} + 8 x - a)

Replace

x

with

u

x = lo g_{10} (- 2 u^{2} + 8 u - a)

Solve for

u, x = lo g_{10} (- 2 u^{2} + 8 u - a)

- \frac{- 8 + 64 + 8 ( - a - 1 0 ^{x} )}{4}, - \frac{- 8 - 64 + 8 ( - a - 1 0 ^{x} )}{4}

f (x) = x^{2} + 18 x + 81

arccos (1)

y = \frac{3 x + 1}{2}

f (x) = \frac{( 3 - 8 x )}{( 4 + 7 x )}

f (x) = \frac{3 x - 2}{7}

What is the inverse of f(x)=log_{10}(-2x^2+8x-a) ?
The inverse of f(x)=log_{10}(-2x^2+8x-a) is -(-8+sqrt(64+8(-a-10^x)))/4 ,-(-8-sqrt(64+8(-a-10^x)))/4