Thread: Least Possible Value of k

1. Least Possible Value of k

If k is an integer and 0.0010101 x 10^(k) is greater than 1,000, what is the LEAST possible value of k?

Set Up:

0.0010101 x 10^(k) > 1,000.

Must I take the log on both sides to find k?

2. Re: Least Possible Value of k

You can eyeball this one as the value of 0 < k will determine how many places the decimal point in the mantissa will move to the right.

3. Re: Least Possible Value of k Originally Posted by harpazo If k is an integer and 0.0010101 x 10^(k) is greater than 1,000, what is the LEAST possible value of k?

Set Up:

0.0010101 x 10^(k) > 1,000.

Must I take the log on both sides to find k?
I don't know if you "must" take the logarithm (especially base 10) of each side to find k, but it is geared for it.

$\displaystyle 0.0010101*10^k > 1,000$

I would isolate the exponential term first:

$\displaystyle 10^k \ > \ \dfrac{1,000}{0.0010101}$

$\displaystyle 10^k \ > 990,000.99 \ \ \ \ \ \ \ \ \$ (rounded to two decimal places)

$\displaystyle \log_{10}(10^k) \ > \ \log_{10}(990,000.99)$

$\displaystyle k \ > \ \log_{10}(990,000.99)$

What is your conclusion from looking at the result after entering the above in a calculator/computer?

4. Re: Least Possible Value of k Originally Posted by greg1313 I don't know if you "must" take the logarithm (especially base 10) of each side to find k, but it is geared for it.

$\displaystyle 0.0010101*10^k > 1,000$

I would isolate the exponential term first:

$\displaystyle 10^k \ > \ \dfrac{1,000}{0.0010101}$

$\displaystyle 10^k \ > 990,000.99 \ \ \ \ \ \ \ \ \$ (rounded to two decimal places)

$\displaystyle \log_{10}(10^k) \ > \ \log_{10}(990,000.99)$

$\displaystyle k \ > \ \log_{10}(990,000.99)$

What is your conclusion from looking at the result after entering the above in a calculator/computer?
My conclusion is that k = 6.

5. Re: Least Possible Value of k Originally Posted by harpazo My conclusion is that k = 6.
Yes, that's what I got from "eyeballing" it as I described. 6. Re: Least Possible Value of k Originally Posted by MarkFL Yes, that's what I got from "eyeballing" it as I described. I needed to work it out to arrive at k = 6.