- Fri May 27, 2016 10:47 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Solve for X (Expressions with Roots)
- Replies:
**3** - Views:
**32**

Solve for x: ux^a = x The answer they give is: Step 1: ux^a = x Step 2: x^(a-1) = (1/u) Step 3: x = (1/u)^((1/(a-1)) I understand until that part. The next step, I don't understand. Step 4: x = u^(1/(1-a)) How do you get from Step 3 to Step 4? They used exponent rules and flipping a subtraction. . ...

- Thu May 12, 2016 6:00 pm
- Forum: Calculus
- Topic: How to prove f(x) = (cx-c*x+x-1)/(2x-x-1), where 1<c<2 is decreasing function?
- Replies:
**3** - Views:
**42**

How to prove the following function is decreasing function? f(x)\, =\, \dfrac{c^x\, -\, cx\, +\, x\, -\, 1}{2^x\, -\, x\, -\, 1} \mbox{with }\, 1\, <\, c\, <\, 2\, \mbox{ and }\, 0.5\, <\, x\, <\, 1 This one probably works just like the other one, so use the same method. What did they use f...

- Thu May 12, 2016 5:46 pm
- Forum: Calculus
- Topic: How to prove f(x) = [(x-1)*cx-2*x*cx-1+x+1]/(2x-x-1) is increasing function?
- Replies:
**1** - Views:
**25**

Can you help me improve the following function f(x) is an increasing function? Sure, but we need to see what you've done first. How far did you get, and where are you stuck? . f(x)\, =\, \dfrac{(x\, -\, 1)\, c^x\, -\, 2xc^{x-1}\, +\, x\, +\, 1}{2^x\, -\, x\, -\, 1} \mbox{ with }\, 1...

- Thu May 12, 2016 5:39 pm
- Forum: Intermediate Algebra
- Topic: Doing a comprehensive review, need fraction help...
- Replies:
**1** - Views:
**33**

The problem is: 2/a 2 - 3/ab + 4/b 2 This is just an expression. Until they give you instructions, there is no problem do resolve. (answer: 2b 2 - 3ab + 4a 2 / a 2 b 2 ) My guess is that they forgot to tell you to "convert to a common denominator and combine". With other fractions that ha...

- Fri Apr 01, 2016 3:09 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: solving for x, given exponentials: 4e^2x + 5e^3x = 15
- Replies:
**7** - Views:
**374**

denicarl wrote:if I have 4e^2x + 5e^3x = 15

how do I solve for x?

What you wrote means this:

a.

Is that right? Or did you mean this?

b.

denicarl wrote:I tried ln 4 +2x+ln5+3x=ln 15 but that doesn't work.

How did you get to this step?

Thanks.

- Fri Apr 01, 2016 3:08 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: solving for x given: 4e^-.2x + 3e^-.3x = 5
- Replies:
**4** - Views:
**124**

denicarl wrote:how do I solve for x given: 4e^-.2x + 3e^-.3x = 5

The way you've written it means this:

a.

Did you mean this?

b.

Thanks.

- Tue Mar 29, 2016 1:03 pm
- Forum: Intermediate Algebra
- Topic: Question about polynomials.
- Replies:
**1** - Views:
**265**

squirrel wrote:Hi all,

I have a quick question regarding polynomials. Can I use long division or synthetic division, or both? Thanks in advance.

Depends on the context, but "both" is probably fine in most cases.

- Sat Jan 23, 2016 1:12 pm
- Forum: Beginning Algebra
- Topic: LCD inquiry
- Replies:
**3** - Views:
**555**

magellan333 wrote:The instructions were to find the LCD of the three. I factored as follows:

X-7 = x-7

3x-21 = 3(x-7)

6x=6x

Therefore my answer was x-7, 3 and 6x.

Okay. Now, after you've read the lesson in the link and factored the 6 in the 6x, what are you getting?

- Wed Jan 20, 2016 11:35 pm
- Forum: Beginning Algebra
- Topic: LCD inquiry
- Replies:
**3** - Views:
**555**

X-7, 3x-21, 6x My answer was 3, x-7, 6x "Your answer" to what? What did the instructions tell you to do? Correct answer per the book is 6x, x-7. If the instructions said to get the factors for the LCD then this answer is right. You can see why here (the last bit of the lesson shows polys;...

- Sat Jan 16, 2016 1:01 am
- Forum: Beginning Algebra
- Topic: How do I know the order (simple I know!)
- Replies:
**1** - Views:
**467**

I have to subtract one equation from another to get my answer. I have to subtract S=0.1M + 18660 from S=5M - 174200. How and why do I know to do S=(5M - 0.1M) - (174200 +18660)? I mean why would I not do '0.1M - 5M'? They said to take 0.1M + 18,600 FROM 5M - 174,200. You're saying you want to do it...