+0

I am probably just missing something.

+2
357
3

This is the problem:

What number does X have to be so that the following equasion is correct:

2^(x-1) + 2^(x-4) + 2^(x-2) = 6,5555...

Nov 27, 2017

#1
+17747
+1

Solve:  2x-1 + 2x-4 + 2x-2  =  6,5555...

First:

6,55555...  =  6 + 5/9  =  54/9 + 5/9  =  59/9

The terms:  2x-1,  2x-4, and 2x-2 can all be written in terms of 2x-4:

2x-1  =  2x-4+3  =  2x-4·23  =  2x-4·8

2x-4  =  2x-4·1

2x-2  =  2x-4+2  =  2x-4·22  =  2x-4·4

Therefore, the problem can be re-written as:

2x-4·8 + 2x-4·1 + 2x-4·4  =  59/9

Factoring out the term  2x-4:

2x-4(8 + 1 + 4)  =  59/9

2x-4(13)  =  59/9

Dividing by 13:

2x-4  =  (59/9) / 13

2x-4  =  59/117

Taking the log of both sides:

log( 2x-4 )  = log(59/117)

Simplifying:

(x-4)·log(2)  =  log(59/117)

Dividing by log(2):

x - 4  =  log(59/117) / log(2)

x  =  log(59/117) / log(2) + 4

x  =  3.012278...

Nov 27, 2017
#2
+1

Solve for x over the real numbers:
2^(x - 4) + 2^(x - 2) + 2^(x - 1) = 59/9

Simplify and substitute y = 2^x.
2^(x - 4) + 2^(x - 2) + 2^(x - 1) = (13×2^x)/16
= (13 y)/16:
(13 y)/16 = 59/9

Multiply both sides by 16/13:
y = 944/117

Substitute back for y = 2^x:
2^x = 944/117

Take the logarithm base 2 of both sides:
x = (ln(944/117))/(ln(2))

Nov 27, 2017
edited by Guest  Nov 27, 2017
#3
+9917
+1

This is the problem:

What number does X have to be so that the following equasion is correct:

2^(x-1) + 2^(x-4) + 2^(x-2) = 6,5555...

Nov 27, 2017