[tex]Given;\ Log_{4}(x^{2}+2x+1)=3 \\ \\ Solution; \\ Log_{4}(x^{2}+2x+1)=3 \\ \\ Log_{4}(x+1)(x+1)=3 \\ \\ Log_{4}(x+1)^{2}=3 \\ \\ 4^{3}=(x+1)^{2} \\ \\ 64=(x+1)^{2} \\ \\ \sqrt{64}=(x+1)^{\not{2}}\ \ \ \ \ \ \ \ \ \ |\ To\ cancel\ the\ exponent,square\ root\ 64... \\ \\ 8=x+1 \\ \\ 8-1=x\ \ \ \ \ \ \ \ \ \ |\ Transpose\ 1... \\ \\ \boxed{\boxed{7=x}}[/tex]
[tex]Checking; \\ Log_{4}(x^{2}+2x+1)=3 \\ \\ Log_{4}(7^{2}+2(7)+1)=3 \\ \\ Log_{4}(49+14+1)=3 \\ \\ Log_{4}(64)=3 \\ \\ 4^{3}=64 \\ \\ 64=64\ \ \ \ \ \ \ \ \ \ |\ \underline{TRUE} \\ \\ \\ Hope\ it\ Helps:) \\ Domini[/tex]
