log question (1 Viewer)

Green Yoda · Jul 3, 2015

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

InteGrand · Jul 3, 2015

Rathin said:
from question
logₐ√x - logₐ1/ₓ
why does ½logₐx -1logₐx
equal to 1.5logₐx

It should be +log_a x, because

$-\log _a \left( \frac{1}{x}\right) = -\left( -\log _a x \right)=\log _a x$ .

Green Yoda · Jul 3, 2015

InteGrand said:
It should be +log_a x, because

$-\log _a \left( \frac{1}{x}\right) = -\left( -\log _a x \right)=\log _a x$ .

oh yes typo my bad.....
but why does 1 log cancel out? to make the answer 3/2log (a) x

InteGrand · Jul 3, 2015

Rathin said:
oh yes typo my bad

So are you essentially asking why $\log _a \frac{1}{x}=-\log _a x$ ?

This is just because $\frac{1}{x}=x^{-1}$ , so $\log _a \frac{1}{x}=\log _a \left( x^{-1}\right)=-\log_a x$ , using the log law $\log _a \left(X^r\right)=r\log _a X$ .

Green Yoda · Jul 3, 2015

InteGrand said:
So are you essentially asking why $\log _a \frac{1}{x}=-\log _a x$ ?

This is just because $\frac{1}{x}=x^{-1}$ , so $\log _a \frac{1}{x}=\log _a \left( x^{-1}\right)=-\log_a x$ , using the log law $\log _a \left(X^r\right)=r\log _a X$ .

no, im asking
why does ½logₐx + 1logₐx
equal to 1.5logₐx

InteGrand · Jul 3, 2015

Rathin said:
no, im asking
why does ½logₐx + 1logₐx
equal to 1.5logₐx

Rathin said:
oh yes typo my bad.....
but why does 1 log cancel out? to make the answer 3/2log (a) x

$What do you mean cancel out? $\frac{1}{2}\log _a x + \log _a x = \frac{3}{2}\log_a x$, because $\frac{1}{2}+1=\frac{3}{2}$. (If it helps, just write $y\equiv \log _a x$, then $\frac{1}{2}\log _a x + \log _a x =\frac{1}{2}y + y=\frac{3}{2}y=\frac{3}{2}\log_a x $. I.e. think of the whole `$\log_a x$' as the pronumeral.)$

Green Yoda · Jul 3, 2015

InteGrand said:
$What do you mean cancel out? $\frac{1}{2}\log _a x + \log _a x = \frac{3}{2}\log_a x$, because $\frac{1}{2}+1=\frac{3}{2}$. (If it helps, just write $y\equiv \log _a x$, then $\frac{1}{2}\log _a x + \log _a x =\frac{1}{2}y + y=\frac{3}{2}y=\frac{3}{2}\log_a x $. I.e. think of the whole `$\log_a x$' as the pronumeral.)$

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 · Jul 3, 2015

Rathin said:
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

kawaiipotato · Jul 3, 2015

Rathin said:
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 · Jul 3, 2015

kawaiipotato said:
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

log question (1 Viewer)

Green Yoda

Hi Φ

InteGrand

Well-Known Member

Green Yoda

Hi Φ

InteGrand

Well-Known Member

Green Yoda

Hi Φ

InteGrand

Well-Known Member

Green Yoda

Hi Φ

Crisium

Pew Pew

kawaiipotato

Well-Known Member

Green Yoda

Hi Φ

Users Who Are Viewing This Thread (Users: 0, Guests: 1)