anonymous
  • anonymous
^3√(2^6)^x I know that the answer is 4^x but what are the steps to solve?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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SolomonZelman
  • SolomonZelman
\(\LARGE \displaystyle \left(\sqrt[3]{2^6}\right)^x\) like this?
anonymous
  • anonymous
parentheses around only the 2^6
SolomonZelman
  • SolomonZelman
\(\LARGE \displaystyle\sqrt[3]{ \left(2^6\right)^x}\)

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anonymous
  • anonymous
yes like that
SolomonZelman
  • SolomonZelman
well, this is what you can do: \(\LARGE \displaystyle\sqrt[3]{ \left(2^6\right)^x}=\left(\sqrt[3]{ 2^6}\right)^x\)
SolomonZelman
  • SolomonZelman
then, deal with what is inside the parenthesis alone, and if the part inside the parenthesis is a 4, then the expression is 4\(^x\).
SolomonZelman
  • SolomonZelman
more percisely. \(\LARGE \displaystyle\sqrt[3]{ \left(2^6\right)^x}=\) \(\LARGE \displaystyle \left(\sqrt[3]{ 2^6}\right)^x=\) \(\LARGE \displaystyle \left(2^{6/3}\right)^x=\) \(\LARGE \displaystyle \left(2^{2}\right)^x=\) \(\LARGE \displaystyle \left(4\right)^x.\)
SolomonZelman
  • SolomonZelman
if you have a question about the validity of the steps (i.e. what propery did I use for a particular step, or why doing something is valid and works, etc) or other questions as wel...
anonymous
  • anonymous
Ok I understand it now... I just wasn't sure about about what to do with the 6 and the 3 but it makes sense now that I see it. Thank you so much!
SolomonZelman
  • SolomonZelman
Anytime..... But, what I did in the first step is not alwyas valid, although is in this case.
SolomonZelman
  • SolomonZelman
For example, if I say: \(\LARGE \sqrt{x^2}=\left( \sqrt{x} \right)^2\) then I am faulty
SolomonZelman
  • SolomonZelman
Why? because the left side is the absolute value of x (i.e. |x| ) And it is continous for all values of x over the interval (-∞,+∞) And the right side is a line y=x for all values of x, where x≥0 (but undefined for x<0)
SolomonZelman
  • SolomonZelman
But, you can always use the following property: \(\LARGE \sqrt[a]{x^b}=x^{b/a}\)
SolomonZelman
  • SolomonZelman
that is what I used to get from line 2 to line 3

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