smurfy14
  • smurfy14
lnx=-2.046 help??
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
If you are trying to solve for x you need to isolate x. To do this here we need to get rid of the ln on the left side...how can we do this?
smurfy14
  • smurfy14
divide
anonymous
  • anonymous
Only problem with division here is that ln is a defined function itself, so dividing it does no good.

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smurfy14
  • smurfy14
so instead you would do e^-2.046 right??
anonymous
  • anonymous
Yes :) You would take e to the power of both sides get what you said on the right, and since e is the inverse of ln, they cancel and form 1.
smurfy14
  • smurfy14
oh ok cool :) thanks! do you think you could help me with one more?
anonymous
  • anonymous
Sure.
smurfy14
  • smurfy14
The temp of an ingot upon being immersed in water kept at 20 degrees Celcius is 20+670e^-5t after t minutes. How long will it take for the temp to reach 25 degrees celcius?
smurfy14
  • smurfy14
for starters, do you know what formula i need to use?
anonymous
  • anonymous
The formula is given. 20+670e^(-5t)=desired temp.
anonymous
  • anonymous
20+670e^(-5t)=25
smurfy14
  • smurfy14
ohh ok
anonymous
  • anonymous
So you need to do the same thing. Isolate t since that is what it is asking for.
smurfy14
  • smurfy14
so first i subtract 20 right?
anonymous
  • anonymous
Yes :)
smurfy14
  • smurfy14
so the next steps would be: ln670+lne^(-57)=ln5 and then would you move the -5t in front of the lne?
anonymous
  • anonymous
You want to take 5 and divide by 670. This will leave you with e^(-5t)=5/670 From here you can do the opposite of last time, which you got right, take the ln of both sides leaving -5t=ln(5/670)
smurfy14
  • smurfy14
oh ok! so i would get -0.992 as the final answer correct?
smurfy14
  • smurfy14
um i dont think i did the last step correctly?
anonymous
  • anonymous
I got about .980 seconds...remember time will never be negative and you can always plug your answer back in to see if it comes out right.
smurfy14
  • smurfy14
ok ok i see what i did wrong thanks! :)
anonymous
  • anonymous
No problem. :)

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