Exponential and Logarithmic Functions - Solve for X

[mysticaurora1]

The question with which I'm requesting assistance in is question 65. It is from the chapter titled "Exponential and Logarithmic Functions" and requires us to solve for X:

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\(\displaystyle \textbf{65. }\, \dfrac{10}{1\, +\, 4\ e^{-0.01x}}\, =\, 2.5\)

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((10/ (1+ 4e^-0.01x) )) = 2.5

Here is my work:

. . . . .\(\displaystyle \dfrac{10}{1\, +\, 4\ e^{-0.01x}}\, =\, 2.5\)

. . . . .\(\displaystyle \dfrac{10}{1}\, +\, \dfrac{10}{4\ e^{-0.01x}}\, =\, 2.5\)

. . . . .\(\displaystyle \dfrac{10}{4\ e^{-0.01x}}\, =\, -7.5\)

. . . . .\(\displaystyle (10)\ (4\ e^{-0.01x})\, =\, -7.5\)

. . . . .\(\displaystyle 4\ e^{-0.01x}\, =\, -0.75\)

. . . . .\(\displaystyle e^{-0.01x}\, =\, -0.1875\)

. . . . .\(\displaystyle \ln\left(e^{-0.01x}\right)\, =\, \ln(-0.1875)\)

. . . . .\(\displaystyle -0.01x\, =\, \ln(-0.1875)\)

. . . . .\(\displaystyle x\, =\, -100\, \ln(-0.1875)\)

I have attached a picture of my work. However; the answer in the textbook states the correct answer is

. . . . .x = -100 ln 3/4

so I don't understand where I've gone wrong. Please guide me towards where I went wrong in my process.

Thanks in advance!

 

[mmm4444bot]

((10/ (1+ 4e^-0.01x) )) = 2.5

… Please guide me towards where I went wrong in my process.

Your first step is incorrect. We may not decompose a ratio by splitting the denominator into pieces.

If 1 + e^(-0.01x) were in the numerator, instead, and 10 in the denominator, then we could decompose the ratio by splitting the numerator into pieces:

(1 + e^[-0.01x])/10 = 1/10 + e^(0.01x)/10

In other words:

(A+B)/C = A/C + B/C

But C/(A+B) is not C/A + C/B

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In your exercise, you need to get x out of the denominator. My first steps would be:

Multiply each side by (1 + e^[-0.01x])

Divide each side by 2.5

Or, if you're familiar with the following pattern, just swap their positions.

A/B = C becomes A/C = B, after multiplying each side by B and dividing each side by C.

PS: The red parentheses highlighted above are unneccessary. :cool:

 

[stapel]

Here is my work:

. . . . .\(\displaystyle \dfrac{10}{1\, +\, 4\ e^{-0.01x}}\, =\, 2.5\)

. . . . .\(\displaystyle \dfrac{10}{1}\, +\, \dfrac{10}{4\ e^{-0.01x}}\, =\, 2.5\)

No; fractions never work this way! Doing the above is like saying the following:

. . . . .\(\displaystyle 1\, =\, \dfrac{2}{2}\, =\, \dfrac{2}{1\, +\, 1}\, \)

. . . . . . . .\(\displaystyle =\, \dfrac{2}{1}\, +\, \dfrac{2}{1}\, =\, 2\, +\, 2\, =\, 4\)

Is 1 equal to 4? No. Neither is 10/(1 + [stuff]) equal to 10 + 10/[stuff].

. . . . .\(\displaystyle \dfrac{10}{4\ e^{-0.01x}}\, =\, -7.5\)

. . . . .\(\displaystyle (10)\ (4\ e^{-0.01x})\, =\, -7.5\)

No; division is not the same as multiplication, and there is no mathematical justification for swapping the operations like this. Doing the above is like saying the following:

. . . . .\(\displaystyle 0.5\, =\, \dfrac{1}{2}\, =\, (1)(2)\, =\, 2\)

Is one-half the same as two? No. Neither is 10/[stuff] equal to 10*[stuff].

. . . . .\(\displaystyle 4\ e^{-0.01x}\, =\, -0.75\)

. . . . .\(\displaystyle e^{-0.01x}\, =\, -0.1875\)

You should have noticed at this step that there was a problem, because e is positive, and e to any power is positive, so there's no way to get a negative result.

. . . . .\(\displaystyle \ln\left(e^{-0.01x}\right)\, =\, \ln(-0.1875)\)

Is it ever possible to take the log of a negative number? No. So the line above should have been a red flag. (Expect to need to look for this on the next test!)

Let's try again, using the rules and properties of logs and exponents that they've given you.

You have a fraction on one side of the equation. The variable is in the denominator of that fraction. You don't want the denominator down below; you want it "upstairs" where you can deal with it. What's the simplest way to accomplish this? (Hint: Read the first reply you received.)

Then you'll have a parenthetical containing the variable on the right-hand side, multiplied by a constant that can go anywhere. You'd prefer it weren't messing with your variable. It's multiplied on where you don't want it. What's the simplest way to move this constant elsewhere? (Hint: Divide.)

You then have two terms on the right-hand side, one of which contains the variable and the other being just a constant that can go anywhere. You'd prefer it weren't messing with your variable. It's currently added on where you don't want it. What's the simplest way to move this constant elsewhere? (Hint: Subtract.)

You now have one term on the right-hand side, containing two factors. One of the factors contains the variable, and the other is just a constant that can go anywhere. You'd prefer it weren't messing with your variable. It's currently multiplied on where you don't want it. What's the simplest way to move this constant elsewhere? (Hint: Divide.)

Do you see now where a "3/4" may be coming from?

Now you have a constant on the left-hand side, and an exponential, all by its lonesome, on the right-hand side. Take logs of either side. What does this give you?

If you get stuck, please reply showing your work in following the above steps. Thank you!