How to x in this eqaution

  • Thread starter nae99
  • Start date
In summary: 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...
  • #1
nae99
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0

Homework Statement



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


Homework Equations





The Attempt at a Solution



4 + 17 (3^2x+1+x) - 7 = 0
 
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  • #2
I am assuming your problem reads:

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
rock.freak667 said:
I am assuming your problem reads:

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.

and how will i expand it...becaues i don't know how to
 
  • #4
nae99 said:
and how will i expand it...becaues i don't know how to

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
rock.freak667 said:
Take the first term 4(32x+1) using the first rule I posted with am+n=aman, how can you rewrite 32x+1 as?

so is it (3^2x)(3^1)
 
  • #6
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
Ivan92 said:
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.

i have no idea what to do next
 
  • #8
32x = (3x)2

& 4(3) = 12

Now, does Ivan's suggestion make more sense to you?
 
  • #9
SammyS said:
32x = (3x)2

& 4(3) = 12

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

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
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
Ivan92 said:
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.

then
12*17(3x)-7=0
 
  • #12
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
Ivan92 said:
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.

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


which would be
36x^2 + 51x - 7=0
 
  • #14
You just did a common algebraic error:

a(bx) [itex]\neq[/itex](ab)x

Just simply put in y for where ever you see 3x. Then do what I told you before.
 
  • #15
Ivan92 said:
You just did a common algebraic error:

a(bx) [itex]\neq[/itex](ab)x

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

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

so are you saying i should do this:

12(y)2+17(y)-7=0
 
  • #16
Ivan92 said:
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

See this equation as corrected by Ivan:

[tex]12(3^x)^2+17(3^x)-7=0[/tex]

Everyplace you see [itex]3^x[/itex] in the equation, replace it with [itex]y\,.[/itex]

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.

Added in Edit:

I see you got the quadratic equation while I was entering my post. I'm a slow typist.
 
  • #17
SammyS said:
See this equation as corrected by Ivan:

[tex]12(3^x)^2+17(3^x)-7=0[/tex]

Everyplace you see [itex]3^x[/itex] in the equation, replace it with [itex]y\,.[/itex]

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.

ok understand now, so i would be

12y^2+17y-7=0
 
  • #18
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
Ivan92 said:
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.

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
Yes, exactly and then use logs to solve it.
 
  • #21
BloodyFrozen said:
Yes, exactly and then use logs to solve it.

so is it
x= log3 1.75
 
  • #22
nae99 said:
i solve foe y but how should i replace it with 3x

for eg. y= 1.75

Not quite.

I get two solutions for y.

One of them is -7/4 which is equal to -1.75.

What's the other solution?
 
  • #23
SammyS said:
Not quite.

I get two solutions for y.

One of them is -7/4 which is equal to -1.75.

What's the other solution?

the other solution is y=0.333
 
  • #24
Actually it's y = 1/3, which is good, because that can be written as 3-1.

So, all you have to do for this solution to y is to solve: [itex]3^x = 3^{-1}\,.[/itex]

Can you solve that for x?
 
  • #25
SammyS said:
Actually it's y = 1/3, which is good, because that can be written as 3-1.

So, all you have to do for this solution to y is to solve: [itex]3^x = 3^{-1}\,.[/itex]

Can you solve that for x?

yes
recall: a^m = a^n, m=n

[itex]3^x = 3^{-1}\,.[/itex]
then x= -1
 

1. How do I solve for x in this equation?

To solve for x in an equation, you must isolate the variable on one side of the equation by performing the same operation on both sides. For example, if the equation is 2x + 5 = 15, you would subtract 5 from both sides, leaving you with 2x = 10. Then, you would divide both sides by 2, giving you the solution of x = 5.

2. What do I do if there are multiple x's in the equation?

If there are multiple x's in the equation, you must first combine like terms on both sides of the equation. Then, follow the same steps as in the previous question to isolate the variable and solve for x.

3. What if the equation contains fractions?

If the equation contains fractions, you will need to multiply both sides by the reciprocal of the fraction. For example, if the equation is (2/3)x = 6, you would multiply both sides by (3/2), leaving you with x = 9.

4. How do I know if I have solved for x correctly?

You can check your solution by plugging it back into the original equation and seeing if it makes the equation true. If it does, then you have solved for x correctly.

5. Is there a specific order in which I must perform operations?

Yes, there is a specific order in which operations must be performed in an equation. This order is known as the order of operations: parentheses, exponents, multiplication and division (performed from left to right), and addition and subtraction (performed from left to right). It is important to follow this order to arrive at the correct solution.

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