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...

Guest Nov 27, 2017

#1**+1 **

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

First:

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

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

2^{x-1} = 2^{x-4+3} = 2^{x-4}·2^{3} = 2^{x-4}·8

2^{x-4} = 2^{x-4}·1

2^{x-2} = 2^{x-4+2} = 2^{x-4}·22 = 2^{x-4}·4

Therefore, the problem can be re-written as:

2^{x-4}·8 + 2^{x-4}·1 + 2^{x-4}·4 = 59/9

Factoring out the term 2^{x-4}:

2^{x-4}(8 + 1 + 4) = 59/9

2^{x-4}(13) = 59/9

Dividing by 13:

2^{x-4} = (59/9) / 13

2^{x-4} = 59/117

Taking the log of both sides:

log( 2^{x-4} ) = log(59/117)

Simplifying:

(x-4)·log(2) = log(59/117)

Dividing by log(2):

x - 4 = log(59/117) / log(2)

Adding 4:

x = log(59/117) / log(2) + 4

x = 3.012278...

geno3141
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))**

Guest Nov 27, 2017

edited by
Guest
Nov 27, 2017