Core 3 Level Help - Solve ln (2y - 1) = 1 + ln (3 - y)

[CathyLou]

Hi.

Could someone please help me with the following question? I would really appreciate any help as I am totally stuck at the moment.

ln (2y - 1) = 1 + ln (3 - y)

Is the answer:

y = (e + 4) / 3 ?

Thank you.

Cathy

 

[Dick]

Nooo. How did you get that?

 

[stewartcs]

Nooo. How did you get that?

Looks like she tried to exponentiate it...which is the incorrect way to solve this problem.

 

[stewartcs]

Hint: Use identities...Ln(x/y) = Ln(x) - Ln(y)

 

[CathyLou]

Nooo. How did you get that?

I went to 2y = 1 = e^1 + (3 - y)

3y = e^1 + 4

y = e^1 + 4 / 3

Could someone please tell me where I went wrong?

Thank you.

Cathy

 

[stewartcs]

Try using the identity that I just posted.

 

[Dick]

e^(1+ln(3-y))=e^1*e^(ln(3-y))=e*(3-y). If you exponentiate a sum, it becomes a product.

 

[CathyLou]

Hint: Use identities...Ln(x/y) = Ln(x) - Ln(y)

Thanks for your help.

I now have y = (3e + 1) / (e + 2)

Cathy

 

[CathyLou]

e^(1+ln(3-y))=e^1*e^(ln(3-y))=e*(3-y). If you exponentiate a sum, it becomes a product.

Thank you for your help!

Cathy

 

[stewartcs]

ln (2y - 1) = 1 + ln (3 - y)

Which gives...

Ln(2y - 1) - Ln(3 - y) = 1

Then using the identity: Ln(x/y) = Ln(x) - Ln(y), gives...

Ln[(2y - 1)/(3 - y)] = 1

Can you take it from here?

 

[stewartcs]

You're welcome.

 

[CathyLou]

Hi.

Could someone please help me with this one too?

Express [(x - 10) / (x - 3)(x + 4)] - [(x - 8) / (x - 3)(2x - 1)] as a single fraction in its simplest form.

I got up to (3x^3 - 20 x^2 + 93x - 126) / (x - 3)(x + 4)(x - 3)(2x - 1) but do not know whether this is correct or what to do next.

Any help would be really appreciated.

Thank you.

Cathy

 

[Dick]

That is correct, but it's unnecessarily complicated. The least common denominator is (x-3)*(x+4)*(2x-1). You don't need the second factor of (x-3).