another simple maths question.

[girlworld_club]

How to solve 9^x - 4.3^x + 3

by using that method where you sub a letter in and solve to find x, sorry for my vagueness.

 

[deswa1]

Assuming the equation was equal to zero.

Let u=3^x -> Therefore u^2-4u+3=0
(u-1)(u-3)=0
u=3, u=1
3^x=1,3
x=0 or x=1

 

[HSC2014]

Whenever you have an unknown power between two terms, look for a common base. Then you can equate them, etc

 

[Leffife]

9^x - 12^x + 3 = 0

Let a = 3^x

3a - 4a + 3 = 0
-a + 3 = 0
a = 3

 

[Amundies]

deswa1, I think in the second line you mean:
"Therefore u^2-4u+3=0"

 

[AAEldar]

9^x - 12^x + 3 = 0

Let a = 3^x

3a - 4a + 3 = 0
-a + 3 = 0
a = 3

Noooo you cannot do this at all. Look at deswa1's post.

 

[Leffife]

Lol yup sorry my bad

 

[Demise]

9^x - 4.3^x + 3

Let u = 3^x

u^2 - 4u + 3 = 0

(u - 3)(u - 1) = 0
u = 3 and u = 1

3^x = 3
ln 3^x = ln 3
x ln 3 = ln 3
x = ln 3 / ln 3
x = 1

3^x = 1
ln 3^x = ln 1
x ln 3 = ln 1
x = ln 1 / ln 3
Since ln 1 = 0, x = 0.

x = 0, 1

 

[Rawf]

Assuming the equation was equal to zero.

Let u=3^x -> Therefore u^2-4x+3=0
(u-1)(u-3)=0
u=3, u=1
3^x=1,3
x=0 or x=1

You accidentally put 4x instead of 4u lolol
It should be u^2 - 4u + 3 = 0

 

[Shadowless]

9^x - 12^x + 3 = 0

Let a = 3^x

3a - 4a + 3 = 0
-a + 3 = 0
a = 3

wtf..

 

[deswa1]

deswa1, I think in the second line you mean:
"Therefore u^2-4u+3=0"

You accidentally put 4x instead of 4u lolol
It should be u^2 - 4u + 3 = 0

Yeah whoops- fixed that

 

[deswa1]

9^x - 4.3^x + 3

Let u = 3^x

u^2 - 4u + 3 = 0

(u - 3)(u - 1) = 0
u = 3 and u = 1

3^x = 3
ln 3^x = ln 3
x ln 3 = ln 3
x = ln 3 / ln 3
x = 1

3^x = 1
ln 3^x = ln 1
x ln 3 = ln 1
x = ln 1 / ln 3
Since ln 1 = 0, x = 0.

x = 0, 1

Please tell me you aren't serious... Yes your working is perfectly correct and the answer is right but when you have something like 3^x=3, you can just write straight out, x=1- its pretty obvious but it'll save you heaps of time in a test.

 

[Demise]

Please tell me you aren't serious... Yes your working is perfectly correct and the answer is right but when you have something like 3^x=3, you can just write straight out, x=1- its pretty obvious but it'll save you heaps of time in a test.

No I'm pretty serious, but the thing is I do that since I freak out in the exam and I end up messing little things up, since I always emphasise the logical flow of mathematical equations. It takes 30 seconds to write and I do it just incase I stuff up.

 

[AAEldar]

No I'm pretty serious, but the thing is I do that since I freak out in the exam and I end up messing little things up, since I always emphasise the logical flow of mathematical equations. It takes 30 seconds to write and I do it just incase I stuff up.

Instead of taking the natural log, take log to the base 3 then (for this question in particular).

 

[deswa1]

No I'm pretty serious, but the thing is I do that since I freak out in the exam and I end up messing little things up, since I always emphasise the logical flow of mathematical equations. It takes 30 seconds to write and I do it just incase I stuff up.

What I would recommend doing is do it quickly like I did it to get x=0,1 and then sub in both values to the original equation to check your answer. This way, you can be sure you are correct.

 

[Demise]

Instead of taking the natural log, take log to the base 3 then (for this question in particular).

That sounds reasonable, good idea. Although in this scenario I used the natural log since it's easily subbed into the calculator, but whatever. Cheers!

 

[girlworld_club]

just one quick question why did you make u= 3^x

this is where my problem is. can you please explain

 

[OzKo]

Because it's a common factor for 9^x and 3^x.

 

[girlworld_club]

But the second part is 4.3^x how did make this into 3^x

 

[deswa1]

just one quick question why did you make u= 3^x

this is where my problem is. can you please explain

You recognise that this will decompose into a quadratic and conventiently 9^x=(3^2)^x=(3^x)^2 -> therefore you can write the function as a quadratic in terms of 3^x