What is the answer to solvefor y,ln(x^2-8y)=x-y-2 ?

The answer to solvefor y,ln(x^2-8y)=x-y-2 is y=\W(-1/(8e^{\frac{x^2-8x+16){8}}})+1/8 x^2

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solvefor y,ln(x^2-8y)=x-y-2

solve for

y, ln (x^{2} - 8 y) = x - y - 2

y = W (- \frac{1}{8 e ^{\frac{x ^{2} - 8 x + 16}{8}}}) + \frac{1}{8} x^{2}

Solution steps

ln (x^{2} - 8 y) = x - y - 2

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

e^{l n (x^{2} - 8 y)} = e^{x - y - 2}

Simplify

e^{l n (x^{2} - 8 y)} : x^{2} - 8 y

x^{2} - 8 y = e^{x - y - 2}

Solve

x^{2} - 8 y = e^{x - y - 2} : y = W (- \frac{1}{8 e ^{\frac{x ^{2} - 8 x + 16}{8}}}) + \frac{1}{8} x^{2}

y = W (- \frac{1}{8 e ^{\frac{x ^{2} - 8 x + 16}{8}}}) + \frac{1}{8} x^{2}

Verify Solutions:

y = W (- \frac{1}{8 e ^{\frac{x ^{2} - 8 x + 16}{8}}}) + \frac{1}{8} x^{2}

True

The solution is

y = W (- \frac{1}{8 e ^{\frac{x ^{2} - 8 x + 16}{8}}}) + \frac{1}{8} x^{2}

lo g_{25} (3 x - 1) = \frac{1}{2}

lo g_{2} (2 x + 3) = 3

lo g_{10} (x) + lo g_{10} (x + 4) = 3

lo g_{4} (5 x + 6) = 0

lo g_{9} (x) + lo g_{9} (8 x - 1) = 1

What is the answer to solvefor y,ln(x^2-8y)=x-y-2 ?
The answer to solvefor y,ln(x^2-8y)=x-y-2 is y=\W(-1/(8e^{\frac{x^2-8x+16){8}}})+1/8 x^2