Hi could somebody please help me answer this question? Thank youuu
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Given:
[tex]\ln(y-8)=-2x+\ln C[/tex]
To find:
y as a function of x.
Solution:
We have,
[tex]\ln(y-8)=-2x+\ln C[/tex]
It can be written as
[tex]\ln(y-8)=\ln e^{-2x}+\ln C[/tex] [tex][\because \ln e^x=x][/tex]
[tex]\ln(y-8)=\ln (Ce^{-2x})[/tex] [tex][\because \ln (ab)=\ln a+\ln b][/tex]
On comparing both sides, we get
[tex]y-8=Ce^{-2x}[/tex]
Adding 8 on both sides, we get
[tex]y=Ce^{-2x}+8[/tex]
Therefore, the equation function is [tex]y=Ce^{-2x}+8[/tex].