Question

log_4(x+4)-log_16x=1
solve

Answer 1

\[\log _{4}(x+4)-\log _{16}x=1\]
\[\log _{4}(x+4)-\frac{\log _{4}x}{\log _{4}4^{2}}=1\]
\[\log _{4}(x+4)-\frac{\log _{4}x}{2}=1\]
\[\log _{4}(x+4)^{2}-\log _{4}x=2\]
\[\log _{4}\frac{(x+4)^{2}}{x}=2\]
\[\frac{(x+4)^2}{x}=4^2\]
\[x^2+8x+16=16x\]
\[x^2+8x=0\]
\[x=0~,~x=-8\]

Answer 2

x^2-8x+16=0
x=4