unceldolan wrote:

Bunuel wrote:

Operations involving the same exponents:Keep the exponent, multiply or divide the bases

\(a^n*b^n=(ab)^n\)

Thus, \(2*2^{k-1}=2^{1+k-1}=2^k\).

For more check here:

math-number-theory-88376.htmlHope it helps.

Hey Karishma, Hey Bunuel,

Till now, I have encountered this kind of problem several times.

Am I right to assume that these are the rules for simplifiying expontents like those in the questions:

\(2^k=2*2^{k-1}\) I can simplify from k to k-1.

\(2^{k+1}=2*2^k\). I can simplify from k+1 to k

BUT I CAN'T simplify the first equation "backwards" meaning that if I see a exponent like \(5^{k-1}\) I have to see directly that I have to get all other exponents to k-1??

I hope you get my question :D Thanks for your help

Greetings!

No, you could also change \(5^{k-1}\) to \(\frac{5^{k}}{5}\)

It is a bit more complicated but may help to understand.

In this case, you would get

\(2^{k}*5^{k-1} = \frac{2^{k} * 5^{k}}{5} = \frac{10^{k}}{5} = \frac{10*10^{k-1}}{5} = 2*10^{k-1}\)

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