Finding the range of the function

[vortex705]

The question is...

If the domain of f(x) = [1/(2^x)] - 7 is all integers, what is a two-digit positive integer that is in the range of f(x)? The answer is either 25 or 57. Can you please tell me how to approach this problem?

I tried to approach this problem graphically, and it's impossible to figure it out that way. Can you help?

 

[arthur ohlsten]

does f[x]=[1/2^x] -7 OR f[x]= 1/[2^x-7]

f[x]= 1/2^x - 7 for 0<x<oo
range is -6 at x=0 to -7 at x=oo
thee range is never positive

f[x]= 1 / [2^x-7] for 0<x<oo
x=0 f[0]= -1/6
x=1 f[1]=-1/5
x=2 f[2]= -1/3
x=3 f[3]=1

for all values of x greater than 2 the range is positive

I don't understand why you give the answer as 25 or 57? Did I goof somehow?

Arthur

 

[arthur ohlsten]

I guess I did goof
Where is 1/[2^x-7] >10
where is 1/10>2^x-7
2^x< 7+1/10
2^x<71/70
x log 2 < log 71 - log 70
x< [log 71-log 70]/ log 2 answer

Arthur

 

[vortex705]

The question is f(x) = (1/(2^x))-7.

The question is a actual practice question from ETS. And the answer given for this question was 25 or 57.

 

[Deleted member 4993]

The question is...

If the domain of f(x) = [1/(2^x)] - 7 is all integers, what is a two-digit positive integer that is in the range of f(x)? The answer is either 25 or 57. Can you please tell me how to approach this problem?

I tried to approach this problem graphically, and it's impossible to figure it out that way. Can you help?

Set of Integers includes negative numbers.

the two digit integers (as power of 2) are 16 (= 9+7), 32(= 25+7) and 64(=57+7).

Now continue....