What is the answer to solvefor y,x=6-ln(2y^3-7) ?

The answer to solvefor y,x=6-ln(2y^3-7) is y=\sqrt[3]{(e^{6-x}+7)/2}

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solvefor y,x=6-ln(2y^3-7)

solve for

y, x = 6 - ln (2 y^{3} - 7)

y = 3 \frac{e ^{6 - x} + 7}{2}

Solution steps

x = 6 - ln (2 y^{3} - 7)

Move

ln (2 y^{3} - 7)

to the left side

x + ln (2 y^{3} - 7) = 6

Move

x

to the right side

ln (2 y^{3} - 7) = 6 - x

Apply log rules

2 y^{3} - 7 = e^{6 - x}

Solve

2 y^{3} - 7 = e^{6 - x} : y = 3 \frac{e ^{6 - x} + 7}{2}

y = 3 \frac{e ^{6 - x} + 7}{2}

Verify Solutions:

y = 3 \frac{e ^{6 - x} + 7}{2}

True

The solution is

y = 3 \frac{e ^{6 - x} + 7}{2}

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

lo g_{5} (x) = 125

\frac{1}{2} lo g_{3} (x - 5) = 2

0 = - 1 - 2 ln (1 - x)

ln (3 x) = \frac{3}{2}

What is the answer to solvefor y,x=6-ln(2y^3-7) ?
The answer to solvefor y,x=6-ln(2y^3-7) is y=\sqrt[3]{(e^{6-x}+7)/2}