Login

Welcome, Guest. Please login or register.

March 29, 2024, 03:25:32 am

Author Topic: How do I solve for x?  (Read 785 times)  Share 

0 Members and 1 Guest are viewing this topic.

Jesse_551

  • Fresh Poster
  • *
  • Posts: 3
  • Respect: 0
How do I solve for x?
« on: October 12, 2019, 06:25:04 pm »
0
This is the question:

Solve 2e^-x + 5 = 3e^-x for x.

It would be much appreciated, thanks.

DrDusk

  • NSW MVP - 2019
  • Forum Leader
  • ****
  • Posts: 504
  • Respect: +130
Re: How do I solve for x?
« Reply #1 on: October 12, 2019, 06:42:05 pm »
+2
This is the question:

Solve 2e^-x + 5 = 3e^-x for x.

It would be much appreciated, thanks.


Jesse_551

  • Fresh Poster
  • *
  • Posts: 3
  • Respect: 0
Re: How do I solve for x?
« Reply #2 on: October 13, 2019, 09:01:28 am »
0



Thank you, can you just clarify steps 2 to 3 I don’t understand how you got from:

e^-x(3-2)=5   to  ln(e^-x)=ln(5)

Thanks again

AngelWings

  • Victorian Moderator
  • ATAR Notes Superstar
  • *****
  • Posts: 2456
  • "Angel wings, please guide me..."
  • Respect: +1425
Re: How do I solve for x?
« Reply #3 on: October 13, 2019, 01:17:14 pm »
+2

Thank you, can you just clarify steps 2 to 3 I don’t understand how you got from:

e^-x(3-2)=5   to  ln(e^-x)=ln(5)

Thanks again
I’m not DrDusk, but I think they did 3 - 2 in the brackets and then used the opposite operation of e, i.e. loge or ln in one step.
VCE: Psych | Eng Lang | LOTE | Methods | Further | Chem                 
Uni: Bachelor of Science (Hons) - genetics
Current: working (sporadically on AN)
VTAC Info Thread

DrDusk

  • NSW MVP - 2019
  • Forum Leader
  • ****
  • Posts: 504
  • Respect: +130
Re: How do I solve for x?
« Reply #4 on: October 13, 2019, 01:48:41 pm »
0

Thank you, can you just clarify steps 2 to 3 I don’t understand how you got from:

e^-x(3-2)=5   to  ln(e^-x)=ln(5)

Thanks again
Yep AngleWings said it.
Because 3-2 = 1 so that because 1*e^-x = 5. Now you take ln which is log base e of both sides which is something you can do and should do as it simplifies any exponential of the form e^(Something).