What is the answer to solvefor x,ln(y)=-2ln(x)+c ?

The answer to solvefor x,ln(y)=-2ln(x)+c is x=e^{-(ln(y)-c)/2}

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solvefor x,ln(y)=-2ln(x)+c

solve for

x, ln (y) = - 2 ln (x) + c

x = e^{- \frac{l n ( y ) - c}{2}}

Solution steps

ln (y) = - 2 ln (x) + c

Switch sides

- 2 ln (x) + c = ln (y)

Move

c

to the right side

- 2 ln (x) = ln (y) - c

Divide both sides by

- 2

ln (x) = - \frac{ln ( y ) - c}{2}

Apply log rules

x = e^{- \frac{l n ( y ) - c}{2}}

12 = - 10 lo g_{10} ((1 - x)^{2})

ln (γ) = ln (4 + e^{2^{β}}) + ω

lo g_{7} (\frac{1}{x}) = - 3

lo g_{2} (x + 3) + lo g_{2} (x - 5) = 0

lo g_{5} (x + 6) = - lo g_{5} (x + 2) + 1

What is the answer to solvefor x,ln(y)=-2ln(x)+c ?
The answer to solvefor x,ln(y)=-2ln(x)+c is x=e^{-(ln(y)-c)/2}