What is the inverse of y=log_{10}(-1/2 x) ?

The inverse of y=log_{10}(-1/2 x) is -5^x*2^{x+1}

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inverse of y=log_{10}(-1/2 x)

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

What is the inverse of y=log_{10}(-1/2 x) ?
The inverse of y=log_{10}(-1/2 x) is -5^x*2^{x+1}