How to go about solving this problem? (3x^2-2x)^6x = 1

Xenon

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Jul 22, 2018
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I'm new here, so hello everybody!

Anyways, I've been struggling a lot with questions where there are variables in the exponent.

How would I solve this problem? (And what would you call a problem like this, so I could try to find more for practice.)

Determine all real numbers x such that (3x2-2x)6x=1


​Thanks!
 

Dr.Peterson

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Nov 12, 2017
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How would I solve this problem? (And what would you call a problem like this, so I could try to find more for practice.)

Determine all real numbers x such that (3x2-2x)6x=1
Ordinarily, you can't use algebraic methods when the variable is in both the base and the exponent. This is a special (which has no name to my knowledge) that requires special thinking.

Just think about what ways there are to get a^b = 1. It might be 1^b, for example, where b could be any real number. What else?

Then solve each of the possible cases. Don't miss any!
 

JeffM

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Sep 14, 2012
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3,244
I'm new here, so hello everybody!

Anyways, I've been struggling a lot with questions where there are variables in the exponent.

How would I solve this problem? (And what would you call a problem like this, so I could try to find more for practice.)

Determine all real numbers x such that (3x2-2x)6x=1


​Thanks!
Like Dr. Peterson, this is a special case within a set of special cases, and I do not know that it has a name.

What I like about the problem is that it encourages you to try something. Here, the obvious thing to try is to apply a log function to both sides of an exponential equation.

\(\displaystyle (3x^2 - 2x)^{6x} = 1 \implies log \left ( \{3x^2 - 2x\}^{6x} \right ) = log(1) \implies 6x * log (3x^2 - 2x) = 0.\)

Do you remember the zero product property?
 

Dr.Peterson

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Nov 12, 2017
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3,098
Like Dr. Peterson, this is a special case within a set of special cases, and I do not know that it has a name.

What I like about the problem is that it encourages you to try something. Here, the obvious thing to try is to apply a log function to both sides of an exponential equation.

\(\displaystyle (3x^2 - 2x)^{6x} = 1 \implies log \left ( \{3x^2 - 2x\}^{6x} \right ) = log(1) \implies 6x * log (3x^2 - 2x) = 0.\)

Do you remember the zero product property?
If you take this approach, just be sure to check whether you have lost any possible solutions (or introduced extraneous solutions) by what you have done. For example, taking the log requires some assumptions about the domain, which may or may not be safe. (I've seen other problems like this that have an extra twist.)

What I like about the problem is that it makes you think, rather than just follow familiar paths.
 

mmm4444bot

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(3x^2-2x)^(6x)
Please note the inserted grouping symbols. :cool:

Without them, your typing could m͏ean: \(\displaystyle (3x^2-2x)^{6} \cdot x\)
 

Denis

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Determine all real numbers x such that (3x2-2x)6x=1
...or get a quadratic:
3x^2 - 2x = 1^(1/(6*x)) : rule: if a^b=c, then a = c^(1/b)
3x^2 - 2x = 1
3x^2 - 2x - 1 = 0
 
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