10 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.The thing inside the square root must be positive, which is true for the denominator portion.
As for the numerator portion, log_0.3(x-1) must be -ve so x must be greater than 2 as x-1>1
Buddy, could you lower the pixel count a bit? It's too good for my device to display. /J
Idk what's happening I'm trying it all but it's just no solving bro ;-;
Toota hua saaz hu mai
To find the domain, the expression inside the square root must be +ve right? The quadratic in denominator is always +ve and its root is also positive. So, the only problematic thing is the -log(0.3)(x-1) in numerator.
Since log is -ve whenever base and argument are on different side of unity, for -log(0.3)(x-1) to be positive, log_0.3(x-1) has to be -ve hence x-1>1 or x>2
Achcha
Ok
Lemme try something then
We need to make different cases for each expression which can be undefined. For instance square root is not always well defined (here you have two of them so two cases possibly) and log is also not always well defined. Make those cases. Solve them. Take intersection of all three because we need all three of them to be true together. Youll be having the domain of definition
Yupps goteem
Thanks gois
+solved @SirLancelotDuLac @Satya S
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