What is the inverse of log_{10}(y/(20)) ?

The inverse of log_{10}(y/(20)) is 2^{y+2}*5^{y+1}

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inverse of log_{10}(y/(20))

inverse

lo g_{10} (\frac{y}{20})

2^{y + 2} \cdot 5^{y + 1}

Solution steps

y = lo g_{10} (\frac{y}{20})

Replace

y

with

y

y = lo g_{10} (\frac{y}{20})

Solve for

y, y = lo g_{10} (\frac{y}{20})

2^{y + 2} \cdot 5^{y + 1}

f (x) = 300 \cdot 4 x

\frac{d}{d} y = 3 x^{2}

f (x) = 5 ln (x) + 2

\frac{8 x - 1}{2 x + 9}

f (x) = - 3^{(x - 7)} + 8.1

What is the inverse of log_{10}(y/(20)) ?
The inverse of log_{10}(y/(20)) is 2^{y+2}*5^{y+1}