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Mathematics 11 Online
Ballery1:

last discontinuity question... :)

Ballery1:

\[f(x) = \frac{ 5 }{ \log_3 x }, x>0\]

Ballery1:

i know that if i plugin x = 3, the log base 3 and the number 3 would cancel out...leaving us with just...f(3) = 5

Ballery1:

it is continuous

JSVSL7:

Set log3(x) to 0. Use the base changing formula. that would be logx/log5 = 0 then logx=0 x=1 so it's discontinuous at 1. also log can never have negative. so it is continuous x>0 and x≠1

Ballery1:

omg good thinking ...i didn't try to think that way... good catch bro :)

JSVSL7:

Kya ya aakhire savaal hai?

Ballery1:

iss exercise/mushk ka akhri swal tha lekin mere pass aur bohot sare sawalaat hein :)

JSVSL7:

theek

Ballery1:

mein zara likh loon bhai...aik menut

JSVSL7:

haan theek

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