logarithms as level help!?

Question

Find the smallest integer x such that 2^-x is less than 10^-6

I did 10^-6 is 1000000
And 2^19 is 524288
and 2^20 is 1048576

So i said that it is 19 as that is under 1000000, but the book says i'm wrong and it's 20. Can anyone explain what i have done wrong?
Thanks x

Answer 1

2^(-x) < 10^(-6)
2^(-x) < 0.000001
lg[2^(-x)] < lg 0.000001
(-x)lg2 < -6
-x < -6/lg2
-x < -19.93
x > 19.93
x = 20

Answer 2

You require 2^(-x) < 10^(-6).
Take logs to base 10:
-x log(2) < -6 log(10)
Since log(10) to base 10 is 1:
-x log(2) < -6
Divide by log(2):
-x < -6 / log(2)
Multiply by -1 (and reverse the inequality):
x > 6 / log(2)
x > 19.93 ...
Therefore since x is to be an integer, x = 20.

Using your technique:
2^(-19) = 1 / 2^19 = 1 / 524288
10^(-6) = 1 / 10^6 = 1 / 1000000
2^(-20) = 1 / 2^20 = 1 /1048576
For 2^(-x) to be less than 10^(-6), 2^x needs to be greater than 10^6.
As x gets bigger, 1/x gets smaller.

Answer 3

Let 2^(-x) = 10^(-6) to find critical value for x.
Take log base ten of both sides.
log means log base ten in the following:-
1 / 2^x = 1 / 10^6
2^x = 10^6
x log 2 = 6 log 10
x log 2 = 6
x = 6 / log 2
x = 20

Answer 4

The result is 20

2^19 =1.9 *10^-6 which is higher than 10^-6

2^-20 =0.95*10^-7

Answer 5

take log of both sides of the inequality
2^x>10^6
x >6 log (base 2) 10

x>6*3,321928095
x>19,93156857

Hence x=20

Answer 6

You are ignoring the fact that the exponent is negative.

2^ -19 is actually still GREATER than 10 ^ -6, because:

1/524288 > 1/ 1000000.

Ok?

Answer 7

Let 2^x = 10^6

Take log_10 of both sides

log_10(2^x) = 6

xlog_10(2) = 6

x = 6 / 0.30 = 20

Answer 8

2^(-x) < 10^(-6)
2^(-x) < 0.000001
lg[2^(-x)] < lg 0.000001
(-x)lg2 < -6
-x < -6/lg2
-x < -19.93
x > 19.93

The smallest integer x is 20.