Using Natural Logarithms



How about you show us whatever start you can make on this one, first.

Please use the ^ symbol and grouping symbols ( ) to be clear.

For example, e0.64t-1 is written as e^(0.64t) - 1

I'm sure you can do the first step.

Next, realize that the denominator is never zero, so you can multiply both sides by it to clear the fraction on the right-hand side.

 
Right on....sorry

50=100(e^0.64t -1/e^0.64t +1)

50/100= (e^0.64t -1/e^0.64t +1)

1/5= (e^0.64t -1/e^0.64t +1 )

e^1/5=(.89648t/2.89648t)

Not sure where to go from here.
 
MathDummie234 said:
50=100(e^0.64t -1/e^0.64t +1)
Click to expand...


You are still not typing these expressions correctly.

50/100 = [e^(0.64t) -1] / [e^(0.64t) + 1]

These parentheses are needed to show which numbers and symbols make up the exponents.

These brackets show clearly what is in the numerator and what is in the denominator of the fraction.

Without these grouping symbols, your typing could mean something else.

-
1/5 = (e^0.64t -1/e^0.64t +1 )
Click to expand...


50/100 does not equal 1/5. :?

Fix that arithmetic mistake, and then multiply both sides of the equation by the denominator. Next, expand out the multiplication that you get on the left-hand side, group the exponential terms, and factor out e^(0.64t).

Isolate this exponential term, and take the natural log of both sides. Solve for t.

If you get stuck, then please show me your work thus far.

 
50=100(e^0.64t -1/e^0.64t +1)

50/100= (e^0.64t -1/e^0.64t +1)

1/2= [(e^0.64t) -1]/[(e^0.64t) +1 )

e^1/2=(.89648t/2.89648t)

2.89648t*e^1/2=(.89648t/2.89648t)*2.....correct?
 
MathDummie234 said:
1/2= [(e^0.64t) -1]/[(e^0.64t) +1 )

Okay -- I can hande this notation, if this is as good as it gets.

e^1/2=(.89648t/2.89648t)

This is not what I meant when I said to factor out the exponential term. It's illegal!

e^(0.64t) is the exponential term.

2.89648t*e^1/2=(.89648t/2.89648t)*2.....correct?

Nope
Click to expand...


I wrote that your next step (after fixing your arithmetic error) is to multiply both sides by the denominator.

Doing this clears the fraction, and it gives you the following equation. (Do you see why?)

\(\displaystyle (e^{0.64t} + 1) \cdot (1/2) = e^{0.64t} - 1\)

Can you multiply out the left-hand side of this equation using the Distributive Property?

Then get the two exponential terms together on one side, and everything else on the other side.

Show me.

 
Sorry to have bothered you this seems more of a burden to help me than an actual help. Thanks for the insight so far I will attempt to stumble my way through the rest of the steps.
 
MathDummie234 said:
… this seems more of a burden to help me than an actual help …
Click to expand...


What do you expect?

Are you looking for something other than the solution steps?

If, instead of guidance, you simply want me to do the work for you, then please say so. I can do that.

 


You're welcome. :)

Hang around. Perhap, Soroban will do this one for you, too.

< Note to self: add walking-on-eggshells course to New Year's Resolutions list >

 
MathDummie234 said:
e^0.64t + 1 = [ (e^0.64t) - 1/1/2 ]

Correct?
Click to expand...


I'm going to continue to be frank. It does not appear, to me, that you have much experience with algebra.

Solving this exercise requires prerequisite skills in manipulating equations algebraically (eg: factoring, combining like terms, multiplying to get rid of parentheses). Without these skills, exercies like this one are going to be torture. Here's the algebra:

\(\displaystyle 50 = 100 \cdot \left( \frac{e^{0.64t} - 1}{e^{0.64t} + 1} \right)\)

\(\displaystyle \frac{50}{100} = \left( \frac{e^{0.64t} - 1}{e^{0.64t} + 1} \right)\)

\(\displaystyle (e^{0.64t} + 1) \cdot \frac{1}{2} = e^{0.64t} - 1\)

\(\displaystyle \frac{1}{2} e^{0.64t} + \frac{1}{2} = e^{0.64t} - 1\)

Subtract (1/2) e^(0.64t) from both sides

Add 1 to both sides

\(\displaystyle e^{0.64t} - \frac{1}{2} e^{0.64t} = \frac{3}{2}\)

Factor out the exponential term

\(\displaystyle e^{0.64t} \cdot (1 - \frac{1}{2}) = \frac{3}{2}\)

\(\displaystyle \frac{1}{2} \cdot e^{0.64t} = \frac{3}{2}\)

Multiply both sides by 2

\(\displaystyle e^{0.64t} = 3\)

Do you know the property of logarithms that allows you to get the variable t out of the exponent position?

All of my algebra to this point was to get the single exponential term equal to a number, so that this particular property could be used to finally solve for t.

Let me know if you need more help with the final few steps. Or, my prior offer to simply finish still stands; all you need to do is ask.
 
Your quite right I teach SERE in the Airforce math is not my cup of tea....I'm actually stationed near you at Fairchild AFB in Spokane and appreciate you taking time out to help the math disabled
 


Now I understand why it's been a long day. You've been under 40 inches of snow for three weeks.

It's more tame on this side of the mountains.

 
Yes it truly sucks! I teach in the northern part of the state where we have gotten around 70 inches with whining students....not a good combination.

Thanks Again,
Now I can use the principles you have taught me to complete the required assignments.

Dwight
 
MathDummie234 said:
Yes it truly sucks! I teach in the northern part of the state where we have gotten around 70 inches with whining students....not a good combination.
Dwight
Click to expand...
Well Dwight, you did a bit of whining yourself at this site :wink:
 
You must log in or register to reply here.
Top