How do you solve 2x−2−x=5?
1 Answer
Jul 8, 2016
Explanation:
Let
Then the equation becomes:
t−1t=5
Multiply through by
0=4t2−20t−4
=(2t−5)2−29
=(2t−5)2−(√29)2
=((2t−5)−√29)((2t−5)+√29)
=(2t−5−√29)(2t−5+√29)
Hence:
2x+1=2⋅2x=2t=5±√29
For any Real value of
So (for Real solutions) we require:
2x+1=5+√29
Hence:
x+1=log2(5+√29)
So:
x=log2(5+√29)−1
To calculate a numerical value for
log2(5+√29)=ln(5+√29)ln(2)≈3.3764522
So:
x≈3.3764522−1=2.3764522