Question

Solve for x without using your calculator:
1. log_3 x² = 2log_3 4 - 3log_3 2
2. 12x³-23x²-3x+2=0

Answer 1

For 1, I have so far:
log_3 x² = -log_3 2
Is that right?

And for 2,
x(12x²-23x-3)-2=0
Do you use the quadratic formula to get two answers, then subtract 2 from both of them with x=0 as a 3rd answer?

Answer 2

For 1, you're almost correct. I'm getting \[\begin{align}\log_3(x^2)&=2\log_3(4)-3\log_3(2) \\
&=\log_3(4^2)-\log_3(2^3) \\
&=\log_3(16)-\log_3(8) \\
&=\log_3(\frac{16}{8})\\
&=\log_3(2)
\end{align}\]

Answer 3

On the left side, you get \[\log_3(x^3)=2\log_3(x).\]Set this equal to the RHS, to get\[\begin{align} 2\log_3(x)&=\log_3(2) \\
\log_3(x)&=\frac{1}{2}\log_3(2)=\log_3(\sqrt2) \\
x&=3^{\log_3(\sqrt2)} \\
x&=\sqrt2
\end{align}\]

Answer 4

Yes okay, I think I got it. :D

Answer 5

Thanks so much!

Answer 6

As for the second one, you'll have to factor it. I wouldn't recommend using the quadratic formula. Give me a minute to see if I can get anywhere with it.

Answer 7

The best way I have for the second one, without a calculator, is guess and check. Which is a terrible method.

Answer 8

However, if we assume this has a nice factorization (one with integers) it has to be of a certain form. Since the leading coefficient is \(12=2^2\cdot3\), we have to have a factorization that looks something like \[(2x \pm ...)(2x\pm...)(3x\pm...)\]or\[(x\pm...)(3x\pm...)(4x\pm...)\]or\[(x\pm...)(x\pm...)(12x\pm...)\]

Answer 9

Let's use \(a, b, c\) for the second numbers in each of these. So we have\[(2x \pm a)(2x\pm b)(3x\pm c)\]\[\vdots\]

Answer 10

So is it just plug in and check until you get it? What if it turns out to not be factorable with integers?

Answer 11

I think I've just come up with a better idea. It's guess and check, but we have limited our guesses. By the rational root theorem, a rational factor must divide \[\frac{p}{q}\]where \(p\) divides 2, and \(q\) divides 12. If you write out some of these, you get that there are only a few possible integer roots. In fact, you would get \(\pm1\) and \(\pm2\). So use guess and check on these.

Start with 1, then 2, then -1, then -2. Once you've found a root this gets much simpler. Is one of these integers a root?

Answer 12

Do you plug 1, -1, 2, and -2 into that factorization outline you made?

Answer 13

No. Into the actual equation. So check f(1), f(2), f(-1), f(-2) where \(f(x)=12x^3-23x^2-3x+2\). If f(x)=0, it's a root.

Answer 14

So is it just 2?

Answer 15

Right, so we have \(x=2\) as one solution. There are still two more. Do you know how to do synthetic division?

Answer 16

I think I remember how.

Answer 17

That would be your next step. Use synthetic division to find \[\frac{12x^3-23x^2-3x+2}{x-2}\]

Answer 18

12x²+x-1 is what I got.

Answer 19

Perfect. Now this, we can factor rather easily. Since we have a \(-1\) at the end, we know that the factorization must look like\[(ax+1)(bx-1)\]So we want two numbers \(a,b\) such that \(a\cdot b=12\), and \(b-a=1\). Can you name two such numbers?

Answer 20

3 and 4

Answer 21

Perfect. That means that \[12x^3-23x^2-3x+2=(x-2)(3x+1)(4x-1)\]Hence, the solutions are \(x=2\), \(x=-\frac{1}{3}\), \(x=\frac{1}{4}\).

Answer 22

Alright, thanks! That was quite a long process. :P

Answer 23

Solving polynomials can be a rather tedious process. Good job though.