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

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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
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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
 
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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.
 

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