Answer:
Assumed that "?????????" means L. And you meant log(x+y)
Let's demonstrate it.
Step-by-step explanation:
[tex]L=\log(x)+\log(1+\frac{y}{x} )\\Let\\\\10^{L}=10^{\log(x)+\log(1+\frac{y}{x} } ) \\\\10^{L}=10^{\log(x)}\times10^{\log(1+\frac{y}{x} })\\10^{L}=x(1+\frac{y}{x})=x+y\\Let\\\log(10^{L})=\log(x+y)\\\\L=\log(x+y)\\\\[/tex]
