# log problem

Printable View

• Jun 23rd 2012, 05:56 AM
mido22
log problem
2 Log(X) + Log 0.1 = Log 5 + Log 2
a a a a

i got X= 10 , and -10
i think all answers accepted because there is power 2 in 2 Log(X)so -10 won't be refused
is that right????
• Jun 23rd 2012, 06:01 AM
johnnylam123
Re: log problem
2 Log(X) + Log 0.1 = Log 5 + Log 2
2logx = log (10/0.1)
2logx = log100
2logx = 2
logx=1
x=10.

x cannot be -10 ,cause logy, y<0 ,is undefined
• Jun 23rd 2012, 06:03 AM
johnnylam123
Re: log problem
note log is a function that have unique sol only if there is
• Jun 23rd 2012, 06:09 AM
mido22
Re: log problem
Quote:

Originally Posted by johnnylam123
2 Log(X) + Log 0.1 = Log 5 + Log 2
2logx = log (10/0.1)
2logx = log100
2logx = 2
logx=1
x=10.

x cannot be -10 ,cause logy, y<0 ,is undefined

ok but 2Log X = Log X^2
so -10 will be 100 and Log 100 = 2???
• Jun 23rd 2012, 06:28 AM
matto
Re: log problem
x=-10 would fail the original equation.
• Jun 23rd 2012, 06:29 AM
mfb
Re: log problem
2Log X = Log X^2 is a formula for positive x only.
In general, $n\, log(|x|) = log(|x|^n)$

With complex numbers, you can evaluate log(-10) (it is a complex number), but you cannot use the formula given above, as -10 is not [real and] positive.
• Jun 23rd 2012, 06:45 AM
Plato
Re: log problem
Quote:

Originally Posted by mido22
2 Log(X) + Log 0.1 = Log 5 + Log 2
a a a a

Is the actual problem $2\log_a(X)+\log_a(0.1)=\log_a(5)+\log_a(2)~?$
• Jun 23rd 2012, 07:59 AM
mido22
Re: log problem
Quote:

Originally Posted by Plato
Is the actual problem $2\log_a(X)+\log_a(0.1)=\log_a(5)+\log_a(2)~?$

no it is without brackets
• Jun 23rd 2012, 08:18 AM
Plato
Re: log problem
Quote:

Originally Posted by mido22
no it is without brackets

What brackets?

Are you talking parenthesis, $(~)~?$.
If you are you need to know that it is incorrect to write $\color{red}\log_a X$.
That is function notation. Just as we write $f(x)$, we also write $\log_a( X).$
• Jun 23rd 2012, 10:12 AM
johnnylam123
Re: log problem
even it is loga, results is the same