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log question (1 Viewer)

Green Yoda

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from question
logₐ√x - logₐ1/ₓ
why does ½logₐx + 1logₐx
equal to 1.5logₐx
 
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Green Yoda

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oh right it helps when you think it as a pronumeral. So technically when 2 logs for example 3 log (x) y + log (x) y, 1 log cancels out to become 3log (x) y? Also is it the same with 3 log (x) y - log (x) y?
 

Crisium

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oh right it helps when you think it as a pronumeral. So technically when 2 logs for example 3 log (x) y + log (x) y, 1 log cancels out to become 3log (x) y? Also is it the same with 3 log (x) y - log (x) y?
3 log (x) y + log (x) y = 4 log (x) y

and

3 log (x) y - log (x) y = 2 log (x) y

You can treat it like a pronumeral it must satisfy BOTH of the following conditions:

* Common Base (in this case x) and

* The Number Following That (in this case y)

If it only satisfies the first condition then you will have to use the multiplication and division log laws
 
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kawaiipotato

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oh right it helps when you think it as a pronumeral. So technically when 2 logs for example 3 log (x) y + log (x) y, 1 log cancels out to become 3log (x) y? Also is it the same with 3 log (x) y - log (x) y?
It's the same as when you have 3x + x = 4x. This case you have 3log(x)y + log(x)y. You add 3 "lots" of log(x)y with 1 "lot" of log(x)y to give 4 "lots" of log(x)y, so 3log(x)y + log(x)y = 4log(x)y
 

Green Yoda

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It's the same as when you have 3x + x = 4x. This case you have 3log(x)y + log(x)y. You add 3 "lots" of log(x)y with 1 "lot" of log(x)y to give 4 "lots" of log(x)y, so 3log(x)y + log(x)y = 4log(x)y
thanks, i fully understand now
 

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