How many integer pairs (x;y) satisfy the equation?

[Violette]

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

 

[blamocur]

You solution looks good to me. I don't know why they list only those numbers in the options since any [imath]x,y[/imath] pair where [imath]y=2x-2[/imath] satisfies the equation. Could there be some additional constraints in there?

P.S. Minor complaint: your use of period for multiplication ("[imath]2.2^{2x+1}[/imath]") looks misleading: most English speakers would treat it as an equivalent of [imath]\left(\frac{22}{10}\right)^{2x+1}[/imath]. Using \cdot for multiplication, i.e. [imath]2\cdot 2^{2x+1}[/imath] would make it more readable.

 

[khansaheb]

Using \cdot for multiplication, i.e. 2⋅22x+12\cdot 2^{2x+1}2⋅22x+1 would make it more readable.

Or use * to indicate multiplication.

 

[Violette]

You solution looks good to me. I don't know why they list only those numbers in the options since any [imath]x,y[/imath] pair where [imath]y=2x-2[/imath] satisfies the equation. Could there be some additional constraints in there?

P.S. Minor complaint: your use of period for multiplication ("[imath]2.2^{2x+1}[/imath]") looks misleading: most English speakers would treat it as an equivalent of [imath]\left(\frac{22}{10}\right)^{2x+1}[/imath]. Using \cdot for multiplication, i.e. [imath]2\cdot 2^{2x+1}[/imath] would make it more readable.

yes I am new to online math typing I am really sorry. They already gives[math]0≤x≤2023;1≤y≤2023[/math] to find how many pairs If i am right but I don't know how to do that ^^.

 

[Dr.Peterson]

You solution looks good to me. I don't know why they list only those numbers in the options since any [imath]x,y[/imath] pair where [imath]y=2x-2[/imath] satisfies the equation. Could there be some additional constraints in there?

Did you miss the part of the problem statement that is only in the title?

How many integer pairs (x;y) satisfy the equation?​

To @Violette: It's a good idea to make sure the entire problem is within the post, rather than splitting it as you did.

So, for how many integers x in [0, 2023] is y = 2x - 2 in [1, 2023]?

 

[blamocur]

How many integer pairs (x;y) satisfy the equation?

Actually, I missed the "how many" part

 

[Violette]

Did you miss the part of the problem statement that is only in the title?

How many integer pairs (x;y) satisfy the equation?​

To @Violette: It's a good idea to make sure the entire problem is within the post, rather than splitting it as you did.

So, for how many integers x in [0, 2023] is y = 2x - 2 in [1, 2023]?

I tried again and let y=2(x-1)[math]=>1≤2(x-1)≤2023<=>\dfrac{3}{2}≤x≤1012.5[/math]and since 1 x produces 1 y so we have 1011 pairs (x;y) ^^