Question

If 10^k = 64, what is the value of 10^(k/2 +1)?

Answer 1

18
42
80
81
320

Answer 2

Ok so it's a matter of breaking it down:

Answer 3

You need to be familiar with the laws of exponents.
\[10^{\frac{k}{2}+1}\]... do you know the law \(a^b*a^c=a^{b+c}\)

Answer 4

yes

Answer 5

so here it's the same idea, but here a=10 and "k/2"=b and "1"=c

Answer 6

can you re-write it using the law?

Answer 7

10^(k/2)+ 10^1

Answer 8

almost... just replace your addition + with a multiplication

Answer 9

10^(k/2) x 10^1

Answer 10

yes, now there is another law \[(a^B)^C=a^{BC}\] and we can apply it to \[10^{\frac{k}{2}}\] by noticing that \[\frac{k}{2}=k*\frac{1}{2}\]

Answer 11

so 320?

Answer 12

Nope =\ If you re-write \[10^{\frac{k}{2}}=10^{k*\frac{1}{2}}\]

Answer 13

You can see that you can say \((10^k)^{\frac{1}{2}}\)

Answer 14

First we notice the following:\[\bf 10^{\frac{k}{2}+1}=10^{\frac{k}{2}}(10)=(10^k)^\frac{1}{2}(10)=10\sqrt{10^k}\]We know that \(\bf 10^k=64\), so plug that in and evaluate.

Answer 15

@sakigirl