Question 397013
This is what's going on (skip this part if you know this already)



{{{2(2^(n-1))}}} Start with the given expression.



{{{2^1(2^(n-1))}}} Rewrite {{{2}}} as {{{2^1}}}



Now we're going to use the identity {{{x^(y)*x^(z)=x^(y+z)}}}



{{{2^(1+n-1)}}} Multiply (by adding exponents) using the identity given above



{{{2^n}}} Add



So {{{2(2^(n-1))=2^n}}}


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Now your question is why isn't {{{2(2^(n-1))}}} equal to {{{4^(n-1)}}}. The simple reason why not is because you can't simply multiply the bases. Remember that {{{2}}} is really {{{2^1}}}



For example, we know that {{{2^3=8}}}. So {{{2*2^3=2*8=16}}}



But if we multiply the bases, then {{{2*2^3=4^3=64}}} which is NOT true.



Also, you CANNOT rewrite {{{2(2^(n-1))}}} into {{{(2^2)^(n-1)}}} because the two are completely different. If you're skeptical about that, plug various values of 'n' into each expression and you'll see different results.



So unfortunately, your line of reasoning was based on a false assumption which lead to that contradiction.



If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=Algebra%20Help">jim_thompson5910@hotmail.com</a>


Also, feel free to check out my <a href="http://www.freewebs.com/jimthompson5910/home.html">tutoring website</a>


Jim