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another simple maths question. (2 Viewers)

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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

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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
 
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HSC2014

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

Leffife

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9^x - 12^x + 3 = 0

Let a = 3^x

3a - 4a + 3 = 0
-a + 3 = 0
a = 3
 

Amundies

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

Demise

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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

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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
 

deswa1

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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

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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

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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

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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

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

this is where my problem is. can you please explain
 

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

deswa1

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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
 

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