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So I'm a high school student and maths never came extremely easily to me. Anyway; I thought I had studied hard for an exam and most questions weren't too hard but one in particular stood out:

simplify the following : $2^{n-1}2^{n-1}4^n$

I got $16^{3n-2}$ but I'm sure that's wrong because I substituted random numbers in and they ended up being different. I understand basic indice and surd laws but I never got around to learning these.

Any help in explaining this would be really helpful. Thanks!

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  • $\begingroup$ No one in the answers told you that $a^n*b^n=(a*b)^n$, you can use it to argue that $2^{n-1}*2^{n-1}=4^{n-1}$. $\endgroup$ – N74 May 30 '17 at 9:24
  • $\begingroup$ try to argue. $2^{n-1}\cdot 2^{n-1} = (2^{n-1})^2 = 2^{2(n-1)} = (2^2)^{n-1}$ $\endgroup$ – John Joy May 31 '17 at 18:04
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    $\begingroup$ I think you have to study some simple books instead of asking questions to get solutions here $\endgroup$ – User-123 Jun 7 '17 at 7:12
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If in doubt, use fractions.

$$2^{n-1}\cdot 2^{n-1}\cdot 4^n=\frac {2^n}2\cdot \frac {2^n}2\cdot 4^n=\frac {4^n}4\cdot {4^n}=\frac {4^{2n}}4=\color{red}{4^{2n-1}}$$

Note that $a^n\cdot a^n=a^{2n}=(a^2)^n$.
Hence $2^n\cdot 2^n=4^n$, and $4^n\cdot 4^n=4^{2n}=16^n$.

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Parentheses, please. What you have written is $2^n-1*2^n-1*4^n=-4^n$. I suspect it is supposed to be $2^{(n-1)}*2^{(n-1)}*4^n$. Now remember the rules of exponents. $a^b \cdot a^c=a^{(b+c)}$ That helps with the first two terms. $(a^b)^c=a^{bc}$ can bring the last term into the mix for you.

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$$2^{n-1}*2^{n-1}*4^n=2^{2(n-1)}*4^n=4^{n-1}*4^n=4^{2n-1}$$ Notice that:

1) $a^b*a^c=a^{b+c}$

2) $a^{(b^ {c})}=(a^b)^c $

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$a^n*a^m = a^{n+m}$ and $(a^n)^m = a^{n*m}$. And that's the entire thing.

So $2^{n-1}*2^{n-1}*4^{n}=$

$2^{n-1}*2^{n-1}*(2^2)^{n}=$

$2^{n-1}*2^{n-1}*2^{2n}=$

$2^{(n-1) + (n-1) + 2n} = 2^{4n -2}$

(which is the same as $4^{2n - 1}$.)

(you can also do this $2^{n-1}*2^{n-1}*4^n = (2^{n-1})^2 *4^n = (2^2)^{n-1}*4^n = 4^{n-1}4^n = 4^{(n-1) + n} = 4^{2n+1}$)

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Ok,Be confident,watch Hall of fame by script :)

If you meant this one like $2^{n-1} * 2^{n-1} * 4^{n}$ Then we could do it like this -

$2^{n-1} * 2^{n-1} * 4^{n} = (2^{n-1})^2 * 4^n $

which is equal to $2^{2n-2}$

where we have used $(a^{b})^{c} = a^{bc}$.

so next $4^{n} = (2^2)^n = 2^{2n}$ (again from the above).

So now we have $2^{n-1} * 2^{n-1} * 4^{n} = (2^{2(n-1)}) * 4^n = (2^{2(n-1)}) * 2^{2n} = 2^{(2(n-1) + 2n)} = 2^{4n-2}$,where we have used $a^{b}.a^{c} = a^{b+c}$.

So we have used two rules - $1) a^{b}*a^{c} = a^{b+c}$ and $2) (a^b)^c = a^{bc}$

In fact you can visualize the 2nd rule from the first as like $(a^{b})^{c} $ is multiplying $a^{b}$ with itself, $c$ times like $a^{b}*a^{b}*...a^{b} $,$c $times, and now you can apply our first rule that is $a^{b+b+b..+b}$, where addition goes $c$ times and thus giving us $a^{bc}$.

I hope it is clear to you now,all the best!

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