Solve 3^(2x) - 2*3^(x+5) + 3^10 = 0 - Brain Boosting Hint

  • Thread starter Thread starter doneitall
  • Start date Start date
  • Tags Tags
    Exponential
Click For Summary
To solve the equation 3^(2x) - 2*3^(x+5) + 3^10 = 0, it can be rewritten as (3^2)^x - 2(3^x)(3^5) + 3^10 = 0, indicating a quadratic form. This allows the use of the Quadratic Formula to find solutions for 3^x. After obtaining the values for 3^x, logarithms can be applied to solve for x. The discussion emphasizes the importance of recognizing the quadratic structure and suggests a substitution method for clarity. The problem is positioned as a review in precalculus, focusing on exponent rules without involving calculus.
doneitall
Messages
3
Reaction score
0
Can someone give me a starter hint for this problem? Brain's not working at full speed...

3^(2x) - 2*3^(x+5) + 3^10 = 0

Thanks
 
Mathematics news on Phys.org
Here are some rules for exponentials:

(a^b)^c=a^{b\cdot c}

a^b\cdot a^c=a^{b+c}

See if you can apply these rules to solve your equation.
 
Welcome to the forum.

Please use the homework outline for your posting of homework.

Are you just trying to solve for x ... or it's derivative or ... what?

Please be specific and show your attempt at a solution and then we can help you out.

Thanks
Matt
 
Looking to solve for x. It is a homework problem... one that I assigned. I teach this for a living and although I'm sure there's something relatively simple that I'm missing, for the life of me I don't see it now. This is a review problem in precalculus so it doesn't involve calculus to solve.

I've looked at it as: (3^2)^x - 2 (3^x)(3^5) + 3^10 = 0 but not sure where to go next. I don't want a solution - just a shove in the right direction. (Before I go mad!)
 
Rewrite your equation as 32x - 2*35*3x + 310 = 0.

This is an equation that is quadratic in form, so you can use the Quadratic Formula to solve for 3x. After that, you can use logs to solve for x.
 
Sheesh! I had tried this by letting u = 3^x but made an error and gave up when it didn't work out. Thanks. I feel better now.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 16 ·
Replies
16
Views
1K
Replies
8
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
3K