Which of the following is equal to 2^k*5^(k-1)?

A. 2*10^(k-1)
B. 5*10^(k-1)
C. 10^k
D. 2*10^k )
E. 10^(2k-1)

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

Answer: A.
_________________

\(2^3 = 2*2*2 = 2*2^2\)

Similarly, \(2^{10} = 2*2^9 = 2^2*2^8\) etc

Hence \(2^k = 2*2^{k-1} = 2^2*2^{k-2} = 2^3*2^{k-3}\) etc

Another Approach: Number Plugging.

Put k = 1 in \(2^k*5^{k-1}\). You get \(2^1*5^0 = 2\)

When you put k = 1 in the options, only option (A) gives you 2.
_________________

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

Is the following also correct ?

2^k x 5^(k-1) = 2^(k) x 5^(k) x 5^(-1)
= 10^(k)/5

?

Nice one


=> 2^k-1 * x 2 x 5^k-1 => 10^k-1 x 2 => option A
_________________

Question is

\(2^k\) \(5^{k-1}\)

[\(2^k\) \(5^k\)] / 5

A) 2 * 10^k * 1/10

[\(2^k\) \(5^k\)] / 5
_________________