Isolate y from equation: `e^y=(1+x^2)*C-1`

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

It is given that `e^y=(1+x^2)*C-1`

To isolate y, take the natural logarithm of both the sides.

=> `ln(e^y) = ln((1 + x^2)*C - 1)`

Use the property `log a^b = b*log a`

=> `y*ln e = ln((1 + x^2)*C - 1)`

The base of natural logarithm is e and `log_b b = 1`

=> `y = ln((1 + x^2)*C - 1)`

The required expression for y is `y = ln((1 + x^2)*C - 1)`

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quantatanu | Student, Undergraduate | (Level 1) Valedictorian

Posted on

Do you want to express something like,  y = function of x, then the solution is given below:

Given expression is:

e^y = (1+x^2)C - 1

Take natural log(log with base e) on both sides:

log [e^y] = log [ (1+x^2 )*C - 1]

=> y = log [ (1+x^2 )*C - 1]

Which is in the desired form.

 

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