How do I solve: 2^1-x = 6 , What is X?

Question

A step-by-step walkthrough would be appreciated

Answer 1

Do you mean

2^(1-x) = 6? take logs, maybe base 2

(1-x) log_2(2) = log_2(6), simplify
(1-x) * 1 = log_2(6), gather, note sign changes

1 - log_2(6) = x.

Done.

Answer 2

Take log base 2 of both sides.
In the following, "log" means log to base 2
(1 - x) log 2 = log 6
1 - x = log 6
x = 1 - log 6

Answer 3

2^1-x = 6

takin log on both sides

(1-x)log2 = log 6

as log2 = 0.3010 and log 6 = 0.7782

(1-x)0.3010 = 0.7782

1-x = 0.7782/0.3010

1-x = 2.5854

-x = 2.5854-1

x= - 1.5854

Answer 4

if it's 2^(1-x)=6, use logarithms
any base would do but i'll use both ln and logb2:
2^(1-x)=6
logb2[2^(1-x)]=logb2(6)
1-x=2.5849625007211561814537389439478
transposing:
x=-1.5849625007211561814537389439478

or:
ln(2^(1-x))=ln(6)
(1-x)ln(2)=ln6
1-x=ln(6)/ln(2)
x=1-ln(6)/ln(2)
x=-1.5849625007211561814537389439478