+118608 Hi Danielr_ddrp
Welcome to the web2.0calc forum 
\(log_2(x+4)-\frac{1}{2}(log_2y)=2\\ log_2(x+4)-log_2(y)^{1/2}=2\\ log_2\frac{x+4}{(y)^{1/2}}=log_24\\ \frac{x+4}{(y)^{1/2}}=4\\ x+4=4(y)^{1/2}\\ x^2+8x+16=16y\\ 16y=x^2+8x+16\)
\(2log_2x-log_2(y-1)=1\\ log_2x^2-log_2(y-1)=log_22\\ log_2\frac{x^2}{y-1}=log_22\\ \frac{x^2}{y-1}=2\\ x^2=2(y-1)\\ x^2=2y-2\\ x^2+2=2y\\ 8(x^2+2)=8*2y\\ 16y=8x^2+16\\\)
SO
\(x^2+8x+16=8x^2+16\\ x^2+8x=8x^2\\ 7x^2-8x=0\\ x(7x-8)=0\\ x=0 \;\;or\;\; x=\frac{8}{7}\\ \)
Log 0 is undefined so x=0 is a nonsense answer.
\(If\;\;x=\frac{8}{7}\;\;\;then\\ 16y=8*\frac{64}{49}+16\\ y=\frac{81}{49}\;\;\;(or\;\;\;1\frac{32}{49})\\ so\;\;x=\frac{8}{7}\;\;\;and\;\; y=\frac{81}{49} \)
I have checked this answer by substituting the values into the original equation.
It is correct :)
+118608 Hi Danielr_ddrp
Welcome to the web2.0calc forum
\(log_2(x+4)-\frac{1}{2}(log_2y)=2\\ log_2(x+4)-log_2(y)^{1/2}=2\\ log_2\frac{x+4}{(y)^{1/2}}=log_24\\ \frac{x+4}{(y)^{1/2}}=4\\ x+4=4(y)^{1/2}\\ x^2+8x+16=16y\\ 16y=x^2+8x+16\)
\(2log_2x-log_2(y-1)=1\\ log_2x^2-log_2(y-1)=log_22\\ log_2\frac{x^2}{y-1}=log_22\\ \frac{x^2}{y-1}=2\\ x^2=2(y-1)\\ x^2=2y-2\\ x^2+2=2y\\ 8(x^2+2)=8*2y\\ 16y=8x^2+16\\\)
SO
\(x^2+8x+16=8x^2+16\\ x^2+8x=8x^2\\ 7x^2-8x=0\\ x(7x-8)=0\\ x=0 \;\;or\;\; x=\frac{8}{7}\\ \)
Log 0 is undefined so x=0 is a nonsense answer.
\(If\;\;x=\frac{8}{7}\;\;\;then\\ 16y=8*\frac{64}{49}+16\\ y=\frac{81}{49}\;\;\;(or\;\;\;1\frac{32}{49})\\ so\;\;x=\frac{8}{7}\;\;\;and\;\; y=\frac{81}{49} \)
I have checked this answer by substituting the values into the original equation.
It is correct :)
