Given that [math]0≤x≤2023;1≤y≤2023[/math] satisfy the equation :
[math]4^{x+1}+log_2(y+3)=2^{y+4}+log_2(2x+1)[/math]My attempt
[math]<=>2.2^{2x+1}-log_2(2x+1)=2.2^{y+3}-log_2(y+3)[/math]both sides share the same structure so i set [math]f(t)=2.2^t-log_2(t)[/math][math]f'(t)=2ln(2)2^t-\dfrac{1}{tln2}>0[/math]so f(t) is a monotonically non-decreasing function(idk if this word is right in my language it's called hàm đồng biến)
[math]=>2x+1=y+3<=>y=2x-2[/math]but here is where i stuck because all 4 options ABCD are A.2023 B.1011 C.2022 D.1012
i don't know what to do next
[math]4^{x+1}+log_2(y+3)=2^{y+4}+log_2(2x+1)[/math]My attempt
[math]<=>2.2^{2x+1}-log_2(2x+1)=2.2^{y+3}-log_2(y+3)[/math]both sides share the same structure so i set [math]f(t)=2.2^t-log_2(t)[/math][math]f'(t)=2ln(2)2^t-\dfrac{1}{tln2}>0[/math]so f(t) is a monotonically non-decreasing function(idk if this word is right in my language it's called hàm đồng biến)
[math]=>2x+1=y+3<=>y=2x-2[/math]but here is where i stuck because all 4 options ABCD are A.2023 B.1011 C.2022 D.1012
i don't know what to do next
Last edited: