Answer:
a = 2 and b = -1.
Step-by-step explanation:
We want to show that the equation:
[tex]\displaystyle \log_{2} \left( y + 1\right) - 1 = 2\log_{2} x[/tex]
Can be rewritten as:
[tex]\displaystyle y = ax^2 + b[/tex]
Where a and b are integers.
Note that:
[tex]\displaystyle 1 = \log _{2} 2[/tex]
Hence:
[tex]\displaystyle \log_{2} \left( y + 1\right) - \log _2 2= 2\log_{2} x[/tex]
From logarithmic properties:
[tex]\displaystyle \log_2 \left(\frac{y+1}{2}\right) = 2\log_2 x[/tex]
Likewise:
[tex]\displaystyle \log_2 \left(\frac{y+1}{2}\right) = \log_2 x^2[/tex]
For the logs to be equivalent, the inner equations must be equivalent. Hence:
[tex]\displaystyle \frac{y+1}{2} = x^2[/tex]
Simplify:
[tex]\displaystyle y = 2x^2 - 1[/tex]
In conclusion, a = 2 and b = -1.
