The range of the equation is [tex]y>2[/tex]
Explanation:
The given equation is [tex]y=2(4)^{x+3}+2[/tex]
We need to determine the range of the equation.
Range:
The range of the function is the set of all dependent y - values for which the function is well defined.
Let us simplify the equation.
Thus, we have;
[tex]y=2 \cdot 4^{x+3}+2[/tex]
This can be written as [tex]y=2^{1+2(x+3)}+2[/tex]
Now, we shall determine the range.
Let us interchange the variables x and y.
Thus, we have;
[tex]x=2^{1+2(y+3)}+2[/tex]
Solving for y, we get;
[tex]x-2=2^{1+2(y+3)}[/tex]
Applying the log rule, if f(x) = g(x) then [tex]\ln (f(x))=\ln (g(x))[/tex], then, we get;
[tex]\ln \left(2^{1+2(y+3)}\right)=\ln (x-2)[/tex]
Simplifying, we get;
[tex](1+2(y+3)) \ln (2)=\ln (x-2)[/tex]
Dividing both sides by [tex]\ln (2)[/tex], we have;
[tex]2 y+7=\frac{\ln (x-2)}{\ln (2)}[/tex]
Subtracting 7 from both sides of the equation, we have;
[tex]2 y=\frac{\ln (x-2)}{\ln (2)}-7[/tex]
Dividing both sides by 2, we get;
[tex]y=\frac{\ln (x-2)-7 \ln (2)}{2 \ln (2)}[/tex]
Let us find the positive values for logs.
Thus, we have,;
[tex]x-2>0[/tex]
[tex]x>2[/tex]
The function domain is [tex]x>2[/tex]
By combining the intervals, the range becomes [tex]y>2[/tex]
Hence, the range of the equation is [tex]y>2[/tex]
