# How to x in this eqaution

1. Jul 7, 2011

### nae99

1. The problem statement, all variables and given/known data

4(3^2x+1) + 17(3^x) - 7 = 0

2. Relevant equations

3. The attempt at a solution

4 + 17 (3^2x+1+x) - 7 = 0

2. Jul 7, 2011

### rock.freak667

4(32x+1) + 17(3x) - 7 = 0

In which case remember that am+n=aman and amn = (am)n.

So after you expand out the first term you can use a substitution to solve for x such as y=3x.

3. Jul 7, 2011

### nae99

and how will i expand it...becaues i dont know how to

4. Jul 7, 2011

### rock.freak667

Take the first term 4(32x+1) using the first rule I posted with am+n=aman, how can you rewrite 32x+1 as?

5. Jul 7, 2011

### nae99

so is it (3^2x)(3^1)

6. Jul 7, 2011

### Ivan92

Yes. So following up on what rock.freak said, you will have

4(3)(32x)+17(3x)-7=0

So now, you will substitute y=3x. After that, you should be able to solve it.

7. Jul 7, 2011

### nae99

i have no idea what to do next

8. Jul 7, 2011

### SammyS

Staff Emeritus
32x = (3x)2

& 4(3) = 12

Now, does Ivan's suggestion make more sense to you?

9. Jul 7, 2011

### nae99

4(3)(3^2x)+17(3x)-7=0

32x = (3x)2

& 4(3) = 12
so after that would it be something like this:

12 + (3^x)^2 + 17(3x)-7=0

10. Jul 7, 2011

### Ivan92

You are not to add 12 but multiply it to the first term. I also noticed that your notation is a bit wrong. Your equation shouldould look like this:

12(3x)2+17(3x)-7=0

Then do what rocky said earlier, which is y=3x. Then you should be able to see what you are to do next.

11. Jul 7, 2011

### nae99

then
12*17(3x)-7=0

12. Jul 7, 2011

### Ivan92

No, you would multiply 12 to the (3x)2.

Where ever you see 3x you substitute in y. When you do that, you should form a quadratic equation. Then do what is necessary to solve it.

12(3x)2+17(3x)-7=0

Here is the equation again.

13. Jul 7, 2011

### nae99

so should it be this
(36x)2 + (51x)-7=0

which would be
36x^2 + 51x - 7=0

14. Jul 7, 2011

### Ivan92

You just did a common algebraic error:

a(bx) $\neq$(ab)x

Just simply put in y for where ever you see 3x. Then do what I told you before.

15. Jul 7, 2011

### nae99

12(3x)2+17(3x)-7=0

so are you saying i should do this:

12(y)2+17(y)-7=0

16. Jul 7, 2011

### SammyS

Staff Emeritus
See this equation as corrected by Ivan:

$$12(3^x)^2+17(3^x)-7=0$$

Everyplace you see $3^x$ in the equation, replace it with $y\,.$

You should obtain a quadratic equation, but the variable is y instead of the usual x which you are more accustomed to.

The expression on the left hand side of your new equation can be factored, which should lead to a solution. Alternatively, you should be able to solve it using the quadratic formula.

I see you got the quadratic equation while I was entering my post. I'm a slow typist.

17. Jul 7, 2011

### nae99

ok understand now, so i would be

12y^2+17y-7=0

18. Jul 7, 2011

### Ivan92

Yep. So now all you would have to do now is solve for y. Then when you solve for y, replace y with 3x then solve for x.

19. Jul 8, 2011

### nae99

i solve foe y but how should i replace it with 3x

for eg. y= 1.75
would it then be
3x=1.75

20. Jul 8, 2011

### BloodyFrozen

Yes, exactly and then use logs to solve it.