What is the answer to solvefor y,ln(2y-5)=x^3+a ?

The answer to solvefor y,ln(2y-5)=x^3+a is y=(e^{x^3+a}+5)/2

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solvefor y,ln(2y-5)=x^3+a

solve for

y, ln (2 y - 5) = x^{3} + a

y = \frac{e ^{x^{3} + a} + 5}{2}

Solution steps

ln (2 y - 5) = x^{3} + a

Apply log rules

2 y - 5 = e^{x^{3} + a}

Solve

2 y - 5 = e^{x^{3} + a} : y = \frac{e ^{x^{3} + a} + 5}{2}

y = \frac{e ^{x^{3} + a} + 5}{2}

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

lo g_{3} (x) - lo g_{3} (2 x - 1) = 0

5 ln (3 x) - 2 ln (2 x) = 6

lo g_{2} (x^{2} + x) = 1

0 = ln (1 + (1 + c^{1}))

What is the answer to solvefor y,ln(2y-5)=x^3+a ?
The answer to solvefor y,ln(2y-5)=x^3+a is y=(e^{x^3+a}+5)/2