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How do you find the y ? ln(y^2-1) - ln(y+1)=ln(sinx).

Answer 1

Anna Allen

I'm assuming you want us to solve for #y#. Use logarithm laws.

#ln((y^2 - 1)/(y +1)) = ln(sinx)#

#ln(((y + 1)(y - 1))/(y +1)) = ln(sinx)#

#ln(y -1) = ln(sinx)#

#y - 1 =sinx#

#y = sinx +1#

However, there will be restrictions on the variable. These will occur when #sinx ≤ 0#.

I hope this is useful!

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Answer 2

Elias Ackerman

To solve the equation ln(y^2 - 1) - ln(y + 1) = ln(sin(x)), first, use the properties of logarithms to combine the terms:

ln((y^2 - 1)/(y + 1)) = ln(sin(x))

Then, use the property of logarithms that states if ln(a) = ln(b), then a = b:

(y^2 - 1)/(y + 1) = sin(x)

Next, cross multiply and rearrange the equation:

y^2 - 1 = (y + 1) * sin(x)

Expand the right side:

y^2 - 1 = y * sin(x) + sin(x)

Now, bring all terms to one side to set the equation to zero:

y^2 - y * sin(x) - sin(x) - 1 = 0

This is a quadratic equation in terms of y. Use the quadratic formula to solve for y:

y = [sin(x) ± sqrt((sin(x))^2 + 4(sin(x) + 1))]/2

These are the solutions for y.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.