Data Sufficiency

This one is driving me crazy...The explanation in the back of the book doesnt help either...

OG11th ed (DS #131)

Is 5^k less than 1000?
(1) 5^k+1 > 3000
(2) 5^k-1 = 5^k-500

_________________________________________________________
My steps
_________________________________________________________
Step 1: Rephrase the question Is 5^k less than 1000?
possible rephrases
a) 5^k < 1000
b) 5^k < (2x5)^3
c) 5^k < 2^3 5^3

Step 2: What underlying concepts can I use to rephrase/understand the question/statements?
USE THE LAW OF INDICES
" a^m x a^n = a^m+n "
" a^-m = 1/a^m "

Step 3: This is where I get lost...can anyone finish my steps and provide an explanation?

_______________________________________________________
Here's how the OG broke the statements in the back of the book...
_______________________________________________________

OG explanation of Statement (1) 5^k+1 > 3000
step 1 - divide both sides of the given inequality by 5 (*see my note*)
5^k+1 ÷ 5 > 3000 ÷ 5 = 5^k > 600

step 2: Statement 1 - INSUFFICIENT
Although 5^k > 600, it is unknown if 5^k < 1000

** The book confuses me here I thought when you divide you're suppose to change the direction of inequality sign **


___________________________________________
OG explanation of Statement (2) 5^k-1 = 5^k-500

step 1- subtract 5^k from both sides
5^k-1 - 5^k = 5^k - 5^k - 500

step 2 - divide all terms by -1
(5^k ÷ -1) - (5^k-1 ÷ -1) = (-500 ÷ -1)
(5^k) - (5^k-1) = (500) **

** The book confuses me here - am I missing something? I thought when you divide all terms by -1, that 5^k will become - 5^k, that -5^k-1 ÷ -1 will be come +5^k-1

step 3 - property of exponents
5^k-5^k (5^-1) = 500

step 4 - substitute for 5^-1
5^k-5^k(1/5) = 500

step 5 - factor out 5^k
5^k(1-1/5) = 500

step 6 - simplify
5^k(4/5) = 500

step 7 - multiply both sides by 5/4
5^k = 500 (5/4)
5^k = 625 which is less than 1000 - SUFFICIENT

____________________________________
This sure seems like a lot of calculation for a data sufficiency problem. I know my other alternative was to pick a number for k, but that seems like a cop out plus its not getting at the root of the concept behind the problem....

Is there an alternative approach to this problem that still applies the concept that is tested ie, the laws of indices?

u shud know 5^1 = 5, 5^2 = 25 5^3 = 125, 5^4 = 625 and 5^5 = 3125
its really easy to calculate for 5

is 5^k < 1000
that means is k < 5

(1) 5^k+1 > 3000

k can be 4/6
Insuff.

(2) 5^k-1 = 5^k-500

the difference between 5^3 and 5^4 is 500

so u can easily determine k = 4

Suff

Choose(b)

All you say is very clear but there are a lot of things to answer and many important points you have to work.

Let's try to solve the question first.

Is 5^k less than 1000?
(1) 5^k+1 > 3000
(2) 5^k-1 = 5^k-500

(1)
5^(k+1)>3000
5^k * 5>3000
You divide by 5 both sides, given 5>0 you do not change anything
5^k>600
As OG explains, this is bigger than 600 but we do not know if it is less than 1000.

(2)
I write you what I did, I don't take the OG solution.
5^k-1 = 5^k-500
I just put 500 to the left and 5^(k-1) to the right consequently I change their sign.
500 = 5^k - 5^(k-1)
I factorize
500 = 5^(k-1)*(5 -1)
500 = 5^(k-1)*4
I divide both sides by 4
125=5^(k-1)

5²=25 5^3=125

So, 5^3=125 and we have 125=5^(k-1)
Then, 3=k-1, then k=4

Now, we can answer the question "Is 5^k less than 1000?"
5^4=625<1000

Hence, B is sufficient

step 1 - divide both sides of the given inequality by 5 (*see my note*)
5^k+1 ÷ 5 > 3000 ÷ 5 = 5^k > 600
** The book confuses me here I thought when you divide you're suppose to change the direction of inequality sign **

When you multiply by a positive number --> Don't change anything
When you multiply by a negative number --> Change the direction of inequality

step 2 - divide all terms by -1
(5^k ÷ -1) - (5^k-1 ÷ -1) = (-500 ÷ -1)
(5^k) - (5^k-1) = (500) **

** The book confuses me here - am I missing something? I thought when you divide all terms by -1, that 5^k will become - 5^k, that -5^k-1 ÷ -1 will be come +5^k-1

Read back what OG tells you, you may confuse something.

Thanks for the helpful replies!