Question

If x=44...4444(100 dights), find the value of log x correct to the nearest integer.

Answer 1

what have you attempted so far?

Answer 2

HINT
44...4444 (100 digits) = 4 * 11...1111 (100 digits)\[
= 4 * (10^{100}+10^{99}+...+10)\]\[
= 4 * 10^{100}(1+0.1+0.01+...)\]

Answer 3

then use standard log rules

Answer 4

For the geometric sum, you might want to simplify using the finite sum formula:
\[\large\sum_{k=1}^nar^{k-1}=\frac{a(1-r^n)}{1-r}\]

Answer 5

Or rather
\[\large\sum_{k=0}^nar^{k}=\frac{a(1-r^{n+1})}{1-r}\]

Answer 6

To the nearest integer? Well if it's log base 10 then this is really just asking if you understand what the log function does.

For instance, what's this? \[\LARGE \log_{10}(100)=2\] what's \[\LARGE \log_{10}(100000)=5\]

So if they tell you it has 100 digits, then you should be able to see where this is going, since log essentially takes a number and gives you approximately how many digits it has.

Answer 7

@Kainui - the question is slightly more involved as it asks for the answer to the nearest integer

Answer 8

That's exactly what I'm addressing my answer to. @asnaseer

Answer 9

I meant it is not just "how many digits are there in the value 444....444"

Answer 10

No, it is. For instance, let's look at the highest possible integer you could put in with 3 digits: and the lowest possible integer you could put in with 3 digits:

http://www.wolframalpha.com/input/?i=log10%28999%29
http://www.wolframalpha.com/input/?i=log10%28100%29

You can see that for a base 10 logarithm the value will always round up to the number of digits, unless it's a power of 10 itself.

Answer 11

ah! - I stand corrected - thanks for clarifying

Answer 12

I don't see anything unacceptable about @Kainui's suggestion. We can confine the value of the logarithm between two integers, since
\[\large\log10^{100}=100~~~~\text{and}~~~~\log10^{101}=101\]
and 444...444 (100 digits) falls in between, so its logarithm will be between 100 and 101. Since 444...444 (100 digits) is less than the midpoint, its logarithm is closer to 100.

Answer 13

yes - I see it now @SithsAndGiggles :)

Answer 14

I guess I was overcomplicating the solution

Answer 15

Your method still checks out though, that's what matters. Log rules would allow you to simplify to
\[\log444...444=\log10^{100}+\log4-\log9\approx100\]