C2 - Logarithms

Right.. and how have you solved them / why do you think its taking so much time? It could be that you're just getting used to it at the start.

Right.. and how have you solved them / why do you think its taking so much time? It could be that you're just getting used to it at the start.

Put up both methods.

Questions: Solve, giving your answer to 3 s.f.

(g) 3^2x+1 - 26(3^x) - 9 = 0
(h) 4(3^2x+1) + 17(3^x) - 7 = 0

How I solved them:

(g)
$3^{2x+1} - 26(3^x) - 9 = 0[br]3^{2x} - 26(3^{x-1}) - 3 = 0[br]3^{2x} - 26 \frac{3^x}{3^1} - 3 = 0[br]Let y = 3^x[br]y^2 - \frac{26y}{3} - 3 = 0[br]3y^2 - 26y - 9 = 0[br]3y^2 + y - 27y - 9 = 0[br]3y(y+1/3) - 27(y+1/3) = 0[br](3y-27)(y+1/3) = 0[br](y-9)(3y+1) = 0[br]y = 9 or y = -1/3[br][br]When y = 9, 3^x = 9$
log₃9 = x
x = 2

When y = -1/3, there are no solutions.

(h)
$4(3^{2x+1}) - 17(3^x) - 7 = 0[br]4(3^{2x}) + 17(3^{x-1}) - 7/3 = 0[br]Let y = 3^x[br]4y^2 - \frac{17y}{3} - 7/3 = 0[br]12y^2 - 17y - 7 = 0[br]12y^2 - 4y + 21y - 7 = 0[br]12y(y-1/3) + 21(y-1/3) = 0[br](12y+21)(y-1/3) = 0[br](4y+7)(3y-1) = 0[br]y = -7/4 or y = 1/3[br][br]When y = 1/3, 3^x = 1/3$
log₃1/3 = x
x = -1

When y = -7/4, there are no solutions.

I see why you feel like it's longer than it should be - you've divided by 3 and then multiplied by 3 again.

When you bring the 3 down in 3^(2x+1), leave it in front, so you get 3*(3^2x) and then substitute as normal. Otherwise, nothing else you can do to speed it up.

(g)

$3^{2x+1} - 26(3^x) - 9 = 0[br]3{(3^x)^2}[sup] [/sup]- 26(3^x) - 9 = 0$
Treat this as a quadratic expression, you can let u=3^x if you want wish to in order for it to appear simpler. Then factorise it.

$(3^x[sup] [/sup]- 9)(3(3^x) + 1) = 0$

Therefore $3^x = 9$ or $-1/3$ ; disregard $-1/3$ as $3^x>0$ for all $x$
-> $3^x=9$ therefore $x=2$ from inspection (I guess you can use logs just to show it's true)

(h)

$4(3^{2x+1}) + 17(3^x) - 7 = 0[br]12{(3^x)^2} + 17(3^x) - 7 = 0$

Treat it as a quadratic again and factorise it.

$(3(3^x) - 1)(4(3^x) + 7) = 0$

Therefore $3^x=1/3$ or $-7/4$ ; disregard negative
-> $3^x=1/3$ therefore $x=-1$ from inspection

The only key thing here is to realise $a^{(b+c)}={a^b}{a^c}$ and $a^{(2b)}=(a^b)^2$

x

x

Does this statement: "You can divide a polynomial by $(x \pm p).$" mean that you can divide a polynomial by (x (a positive or negative number) so e.g. (x-2) or (x+2) right? I feel I misinterpreted it when I thought it meant you can divide any polynomial by p as a negative or positive number so again (x-2) and (x+2).

I know that you probably won't get what I'm saying lool, but I hope you do.

I'm not entirely sure what the difference is between what you've said and what they've said.

If you fix a p value, whether it's positive or negative, the + / - will cover both ends.

But you're getting far too bogged down in the specification and such statements what's important is that you can divide by (x+a) or (x-a) where a is a positive constant.

Lol, no what I was thinking was:

If a polynomial is let's say... x^3 - 10x^2 + 19x + 30, since (x+1) is a factor, I was thinking that if p = 1, (x-1) is also a factor. But, that is incorrect because if you factorise it, you get: (x+1)(x-5)(x-6) lol.
I hope you understand what I am saying now .

Does this statement: "You can divide a polynomial by $(x \pm p).$" mean that you can divide a polynomial by (x (a positive or negative number) so e.g. (x-2) or (x+2) right? I feel I misinterpreted it when I thought it meant you can divide any polynomial by p as a negative or positive number so again (x-2) and (x+2).I know that you probably won't get what I'm saying lool, but I hope you do.

Lol, no what I was thinking was:If a polynomial is let's say... x^3 - 10x^2 + 19x + 30, since (x+1) is a factor, I was thinking that if p = 1, (x-1) is also a factor. But, that is incorrect because if you factorise it, you get: (x+1)(x-5)(x-6) lol.I hope you understand what I am saying now .

Ohh, I see what you mean. but yeah - just go with what I said about 'a' in my previous post.

I think I know what you mean and I see the flaw in your thought. Let's say you have some polynomial $p(x)$.

The book is basically saying you CAN DIVIDE the polynomial by any linear expression such as $(x \pm a)$ however this does NOT mean $x=a$ and $x=-a$ are BOTH roots of $p(x)=0$, therefore it does NOT mean $(x-a)$ and $(x+a)$ are both factors of $p(x)$. ONLY if $\frac{p(x)}{(x-a)}$ gives a remainder of 0 will $x=a$ be a factor of it. (or $x=-a$ if divided through by $(x+a)$)

Just because you can divide a polynomial by a linear expression, it does not strictly mean that linear expression has a root for the polynomial.

Practice long division with polynomial expressions to see what I mean.

Just realised p(n) should be written as p(x) instead. Wouldn't make sense for a function of p to be expressed in terms of x, sorry. xD
But this is what I mean, therefore x=2 a real root if p(x)=0. If you plug in x=-2, the first division will not be determined because you would be dividing by 0 on the denominator, thus x=-2 cannot be a root.



Don't worry, I've done a lot of long divsion :P. Loads of questions and practice, mastered it. I just got a bit confused on a different question and realised this, hence why I asked. Thanks for the explanation though .
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In all honesty, I knew where I was wrong and what it meant. I just wanted to see you how both would react to it and explain it to me !



Yeah , I realised that haha. Thanks for explaining it .

Well lucky you I was bored enough to explain it detail, otherwise I would've left SeanFM's explanation for you :P

In all honesty, I knew where I was wrong and what it meant. I just wanted to see you how both would react to it and explain it to me !

OP please do me a favour and solve x²+2x+2=0 , I'm so stuck on it!

Riiight. Thanks for that ...