Exponential Growth for Pre-Calculus

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PrecalcStuden
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Hi everyone.

My final is coming up for Precalc, and I'm studying my butt off.

I was really needing help with this Exponential Growth Equation to find variable t.

8.0e^(.033t) = 59.8e^(.001t)

(8 times e to the .033t equals 59.8 times e to the .001t)

I would greatly appreciate this because I'm stressing out for my finals. I have always been bad at this stuff
 
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CRGreathouse said:
Just take the logarithm of both sides and it's easy to solve.

I know it sounds like I'm asking you to do it for me, but I really don't know where to begin. I'm confuzed because of the coefficients, because i can just natural log both sides if there weren't any.

If you can please guide me good sir, I KNOW I can do the rest of this subchapter by myself. Thanks a bunch
 
CRGreathouse said:
Just take the logarithm of both sides and it's easy to solve.

I know it's simple for you, but I swear it's so complicated for me. Please, this would really help me my kind sir
 
Integral said:
What do you know about logs?

Review the basic laws, look especially at logs and multiplication. Also study the relationship between e and ln.

I know about logs my good sir. It's just this one problem tripping me up. I just really need someone to show me how its done, and your help will defaintely be appreciated. Should i make one side equal zero? i just don't know,
 
will someone please help, this is the last problem i need and I am done with this sub chater
 
Please help me with this 1 equation. Then i'll be done with this subchapter

Please, I really need help with this.

I have to solve for t here.

8.0e^( .033t ) = 59.8e^ (.001t)

Please, i just really need you to show me how this is done.
 


Borek said:
exey = ex+y

I know that, but this is a weird one with weird coefficients. Please just walk me through this. gosh, I am stressing out already
 
CRGreathouse said:
Just take the logarithm of both sides and it's easy to solve.
So CRGreathouse is suggesting to do this:

[tex]8.0e^{.033t} = 59.8e^{.001t}[/tex]

[tex]\ln{(8.0e^{.033t})} = \ln{(59.8e^{.001t})}[/tex]

Tell us what to do next.01
 
so will it then turn into

ln 8.0 (.033t) = ln 59.8 (.001t)

Please I've been wait all day just to solve this one dang problem.
 
Integral said:
No, You need to use the laws for logs and multiplication.

integral, i beg of you. i just really need someone to help me with this mroe in depth. I've been waiting all day, i would really appreciate it integral. please
 
would someone pklease help me. I am begign you guys. I need help with this. geeze
 


would someone please help me I am freaking begging you. I am down on my knees
 
WOULD SOMEONE PLEASE HELP ME

im begging yiou guys
 
We need to see some effort on your part. Show me that you have even tried to apply the hints you have been given.

Your problem is of the form:

[tex]A e^x = B e^y[/tex]
So taking the ln:
[tex]ln(A e^x) = ln (B e^y)[/tex]

since ln(a * b ) = lna + lnb
[tex]lnA + ln(e^x) = lnB + ln(e^y)[/tex]

Can you finish?
 
Integral said:
We need to see some effort on your part. Show me that you have even tried to apply the hints you have been given.

Your problem is of the form:

[tex]A e^x = B e^y[/tex]
So taking the ln:
[tex]ln(A e^x) = ln (B e^y)[/tex]

since ln(a * b ) = lna + lnb
[tex]lnA + ln(e^x) = lnB + ln(e^y)[/tex]

Can you finish?

Integral, thank you kind sir.

Let's see here.

Yes sir.

Looks like it will be

ln 80 + .033t = ln 59.8 + .001t

From there

ln 80 - ln 59.8 = .001t - .033t

ln 80 - ln 59.8 = -.032
then divide and use a calculatrosomething isn't adding up quite right sir, I am getting a negattive numbe[/I]

wait, edit, it's ln 8 not 80. Thank you integral YOU ARE SO NICE. thank u so much.
 
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