# Thread: Factoring to reduce Complex Fractions. Exponents laws.

1. ## Factoring to reduce Complex Fractions. Exponents laws.

It's a bit odd...how I am in Math 3 and still have trouble with factoring! (Please don't tell my math teacher...she won't take it too kindly...)

"Express in simplest form"
$\displaystyle \frac{a^2-9}{a^2-3a} \bullet \frac{a^2+a}{a+3}$
Work done so far: (Unsure about accuracy...)
$\displaystyle \frac{(a+3)(a-3)}{(a+1)(a-3a)}\bullet\frac{(a+1)(a+a)}{(a+3)}$$\displaystyle = \frac{(a-3)}{(a-3a)}\bullet\frac{(a+a)}{1} "Solve for x: \displaystyle 2^x+2 = 4^x-1" Don't even know how to begin this one...set it equal to 0? 2. ## Difference of squares; Undistribute. "Express in simplest form" ---------------------------------------- Difference of squares, and undistribution: [(a+3)(a-3)(a)(a+1)] / [(a)(a-3)(a+3)] Cancellation: a+1 is what remains, I think. Take a look, I will return soon. 3. "Solve for x: " Don't even know how to begin this one...set it equal to 0 x log 2 + log 2 = x log 4 - log 1 or this, if you like: 2^x - 4^x = -3 x log 2 - x log 4 = log (-3) x (log 2 - log 4) = log (-3) Take it from there on your scientific calculator. I shall return soon... 4. ## No more left... I thought this mess was another problem but I see now that it is part of a solution: ------------------------------------------------ Work done so far: (Unsure about accuracy...) \underline{(a+3)(a-3)}(a+1)(a-3a) \bullet \underline{a+1)(a+a)}(a+3) = \underline{(a-3)}(a-3a) \bullet \underline{(a+a)}1 ------------------------------------------------ I don't use LaTex or whatever this mess started out as. What is Math 3? Looks a lot like what was called Algebra 2 where I went to school. Bye. 5. Thanks for your help Bradley! Yes, I am using LaTex. Math 3 is an integrated math course focusing on Algebra II, Geometry, and some Trig. (Basically)... Any further help is appreciated! 6. Originally Posted by MrOats It's a bit odd...how I am in Math 3 and still have trouble with factoring! (Please don't tell my math teacher...she won't take it too kindly...) "Express in simplest form" \displaystyle \frac{a^2-9}{a^2-3a} \bullet \frac{a^2+a}{a+3} Work done so far: (Unsure about accuracy...) \displaystyle \frac{(a+3)(a-3)}{(a+1)(a-3a)}\bullet\frac{(a+1)(a+a)}{(a+3)}$$\displaystyle = \frac{(a-3)}{(a-3a)}\bullet\frac{(a+a)}{1}$
"Solve for x: $\displaystyle 2^x+2 = 4^x-1$"
Don't even know how to begin this one...set it equal to 0?
$\displaystyle \frac{a^2-9}{a^2-3a}\cdot\frac{a^2+a}{a+3} \implies \frac{(a+3)(a-3)}{a(a-3)}\cdot\frac{a(a+1)}{a+3}\implies a+1$

For the second question, I will think outside the box:

Since $\displaystyle 4=2^2$, then

$\displaystyle 2^x+2=2^{2x}-1\implies 2^{2x}-2^x-3=0$

Making a substitution $\displaystyle u=2^x$, we get the quadratic:

$\displaystyle u^2-u-3=0 \implies u=\frac{1\pm\sqrt{1+12}}{2}$.

Thus,

$\displaystyle 2^x=\frac{1\pm\sqrt{1+12}}{2}$. We ignore the negative answer, because $\displaystyle 2^x$ can NEVER be negative.

Thus, $\displaystyle 2^x=\frac{1+\sqrt{13}}{2} \implies x=\frac{1+\sqrt{13}}{2\log 2}$

Hope this helped you out!!

"Solve for x: "
Don't even know how to begin this one...set it equal to 0

x log 2 + log 2 = x log 4 - log 1
or this, if you like:

2^x - 4^x = -3
x log 2 - x log 4 = $\displaystyle {\color{red}log (-3)}$
x (log 2 - log 4) = $\displaystyle {\color{red}log (-3)}$
Take it from there on your scientific calculator.

I shall return soon...
Can't take the log of something negative!!

8. THANK YOU, Chris, for the informative and extensive, detailed help!

9. Originally Posted by Chris L T521
$\displaystyle \frac{a^2-9}{a^2-3a}\cdot\frac{a^2+a}{a+3} \implies \frac{(a+3)(a-3)}{a(a-3)}\cdot\frac{a(a+1)}{a+3}\implies a+1$

For the second question, I will think outside the box:

Since $\displaystyle 4=2^2$, then

$\displaystyle 2^x+2=2^{2x}-1\implies 2^{2x}-2^x-3=0$

Making a substitution $\displaystyle u=2^x$, we get the quadratic:

$\displaystyle u^2-u-3=0 \implies u=\frac{1\pm\sqrt{1+12}}{2}$.

Thus,

$\displaystyle 2^x=\frac{1\pm\sqrt{1+12}}{2}$. We ignore the negative answer, because $\displaystyle 2^x$ can NEVER be negative.

Thus, $\displaystyle 2^x=\frac{1+\sqrt{13}}{2} \implies x=\frac{1+\sqrt{13}}{2\log 2}$

Hope this helped you out!!
I caught a typo on my part!

$\displaystyle 2^x=\frac{1+\sqrt{13}}{2}\implies x\log2=\log\left(\frac{1+\sqrt{13}}{2}\right)\impl ies x=\frac{\log\left(\frac{1+\sqrt{13}}{2}\right)}{\l og2}\approx 1.20337$