# Thread: Help with gr 12 math?

1. ## Help with gr 12 math?

How would I do this question below:

11. Solve and check for extraneous roots. Where necessary, round answers to 2 decimal places.

$\displaystyle 3^2x = 7(3^x)-12$

The 2x is the power of 3.

I have no idea how to even start that. I tried dividing it by 7 first. Didn't work. Then I tried bringing over the 12 and that didn't work either.

2. $\displaystyle 3^{2x} = 7(3^x)-12$

$\displaystyle (3^x)^2 = 7(3^x)-12$

make $\displaystyle 3^x = a$

$\displaystyle a^2 = 7a-12$

$\displaystyle a^2 - 7a+12=0$

$\displaystyle (a-3)(a-4)=0$

$\displaystyle a=3,4$

$\displaystyle 3^x=3,4$

$\displaystyle x=1,\log_34$

3. Originally Posted by pickslides
$\displaystyle 3^{2x} = 7(3^x)-12$

$\displaystyle (3^x)^2 = 7(3^x)-12$

make $\displaystyle 3^x = a$

$\displaystyle a^2 = 7a-12$

$\displaystyle a^2 - 7a+12=0$

$\displaystyle (a-3)(a-4)=0$

$\displaystyle a=3,4$

$\displaystyle 3^x=3,4$

$\displaystyle x=1,\log_34$
thank you. but how did you get $\displaystyle 3^{2x}$ to letting a be $\displaystyle a^2$ when it was $\displaystyle 2x$ and not $\displaystyle x^2$?

4. $\displaystyle 3^{2x} = (3^x)^2$

comes from the index law

$\displaystyle (a^m)^n = a^{m\times n}$

5. Originally Posted by pickslides
$\displaystyle 3^{2x} = (3^x)^2$

comes from the index law

$\displaystyle (a^m)^n = a^{m\times n}$
thank you. So when I applied when I learned from you to my next question I got it wrong. This is what I did.

Solve and check for extraneous roots.

$\displaystyle 5^x = 3 - 5^{2x}$

Let a be $\displaystyle 5^x$

$\displaystyle a = 3 - a^2$
$\displaystyle a^2+a-3$
$\displaystyle a = -2.23$ or $\displaystyle 1.232$

The answer in the book is a = 0.16

6. Originally Posted by Barthayn

$\displaystyle a = 3 - a^2$
$\displaystyle a^2+a-3$
$\displaystyle a = -2.23$ or $\displaystyle 1.232$
assuming your solutions for a are correct you must then re-visit what you made a

$\displaystyle 5^x = -2.23$ or $\displaystyle 5^x= 1.232$

The first cannot be solved because you can't have the log of a negative number so you have only one solution $\displaystyle x= \log_51.232$

7. according to the answer book the answer is 0.16. What did I do wrong?

8. Originally Posted by Barthayn
according to the answer book the answer is 0.16. What did I do wrong?
Using the quadratic equation, I found roots to be:

1.303
-2.303

So, when solving $\displaystyle a^2 + a -3$, something went awry.

What pickslides said about negative logs holds true here, so -2.303 isn't a solution.

When you solve for x

$\displaystyle 5^x = 1.303$
using logs, x = .1644

which is probably what the book is talking about.

9. Maybe this is a longwinded way to do it:

$\displaystyle 5^x=3-5^{2x}$

let $\displaystyle 5^x=a$

$\displaystyle a^2+a-3=0$

$\displaystyle \left (a+\frac{1+\sqrt{13}}{2}\right )\left (a+\frac{1-\sqrt{13}}{2}\right )=0$

$\displaystyle 5^x=\frac{\sqrt{13}-1}{2}$ (as previously stated by Pickslides, you can disregard the other solution, since you can't take the log of a negative value)

$\displaystyle x=\frac{log(\frac{1}{2}(\sqrt{13}-1))}{log5}$

$\displaystyle x\approx 0.16$

10. Originally Posted by pickslides
assuming your solutions for a are correct you must then re-visit what you made a

$\displaystyle 5^x = -2.23$ or $\displaystyle 5^x= 1.232$

The first cannot be solved because you can't have the log of a negative number so you have only one solution $\displaystyle x= \log_51.232$
Originally Posted by Stroodle
Maybe this is a longwinded way to do it:

$\displaystyle 5^x=3-5^{2x}$

let $\displaystyle 5^x=a$

$\displaystyle a^2+a-3=0$

$\displaystyle \left (a+\frac{1+\sqrt{13}}{2}\right )\left (a+\frac{1-\sqrt{13}}{2}\right )=0$

$\displaystyle 5^x=\frac{\sqrt{13}-1}{2}$ (as previously stated by Pickslides, you can disregard the other solution, since you can't take the log of a negative value)

$\displaystyle x=\frac{log(\frac{1}{2}(\sqrt{13}-1))}{log5}$

$\displaystyle x\approx 0.16$
thank you all. First I did not add the 1 to the 12 as well then log know variable to get x by its self.