What is the inverse of-4log_{10}(-1/3 (x+2))+1 ?

The inverse of-4log_{10}(-1/3 (x+2))+1 is -3*10^{-(x-1)/4}-2

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inverse of-4log_{10}(-1/3 (x+2))+1

inverse

- 4 lo g_{10} (- \frac{1}{3} (x + 2)) + 1

- 3 \cdot 1 0^{- \frac{x - 1}{4}} - 2

Solution steps

y = - 4 lo g_{10} (- \frac{1}{3} (x + 2)) + 1

Replace

x

with

y

x = - 4 lo g_{10} (- \frac{1}{3} (y + 2)) + 1

Solve for

y, x = - 4 lo g_{10} (- \frac{1}{3} (y + 2)) + 1

- 3 \cdot 1 0^{- \frac{x - 1}{4}} - 2

x^{2} - 3 x - 10

f (x) = - (x - 1) - 1

- x^{2} - 2 x

y = - 2 \cdot (\frac{1}{3})^{x}

\frac{4 - x ^{2}}{x + 4}

What is the inverse of-4log_{10}(-1/3 (x+2))+1 ?
The inverse of-4log_{10}(-1/3 (x+2))+1 is -3*10^{-(x-1)/4}-2