Pre-Calc using log/ln: (2^x) +(2^-x) = 6
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mmm4444bot
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I agree that the Quadratic Formula will be helpful, but that's not something you can see until after you've applied some algebra.
Do you remember the definition of a negative exponent? If you apply that definition, you'll see that x appears somewhere in a denominator. :idea: In general, we cannot solve for x while it appears in a denominator.
Use algebra, to get an equivalent equation where no x appears in any denominator.
Please show your work, up to that point, and we can go from there.
If you don't remember the meaning of negative exponents, look it up.
After that, if you can't think of any algebraic steps to get x out of the denominator, try making the following substitution, and see if it gives you an idea.
Let u = 2^x
PS: Please also read the forum guidelines. Thank you! :cool:
Do you remember the definition of a negative exponent? If you apply that definition, you'll see that x appears somewhere in a denominator. :idea: In general, we cannot solve for x while it appears in a denominator.
Use algebra, to get an equivalent equation where no x appears in any denominator.
Please show your work, up to that point, and we can go from there.
If you don't remember the meaning of negative exponents, look it up.
After that, if you can't think of any algebraic steps to get x out of the denominator, try making the following substitution, and see if it gives you an idea.
Let u = 2^x
PS: Please also read the forum guidelines. Thank you! :cool:
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