another gchem/math combination q

[113zami]

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yet another one of those gchem/math combined q's

4.4909=4.55-(.0591/3)Log(10^3/x)

x is supposed to be the molarity of Au and I need to solve for x, ansr says x=1M but I keep getting -1, I know molarity should be positive but I don't know why i am getting (-) ansr (probably another careless mistake that I couldn't figure out), this is how I did it:

4.4909-4.55= -(.0591/3)Log(10^3/x)
-.0591(-3/.0591)=Log(10^3/x)
3=Log(10^3)-Logx
3=3-Logx
3-3=-Logx
0=-Logx
-1=x

please help, thanks

 

[jntruong2003]

hey 13zami,

when you are at 3=log10^3/x, you should realize that log10=1.
so 3=(3/x)log10 become 3=3/x.
then x=1.
hope that helps.

 

[113zami]

oh, my bad, forgot () there, the 3=log10^3/x should actually be 3=log10^(3)/x so it's 3=log1000/x so the 10 is only raised to the 3 not the x
so how would the asr be

 

[jntruong2003]

so are you saying it is 3=(log10^3)/x ?
regardless if the problem is 3=log10^(3)/x, the problem will still become
3=(3)/xlog10
3=(3)/x(1) (log10=1)
x=1

 

[113zami]

you totally lost me there, first, i don't know if it will make a difference but the way its written is that only the 10 to the power 3 is divided by x, so log is not divide by x, i think i missed another (), so that's why I used this property logx/y=logx-logy, which means 3=log(10^(3)/x) (i think that's how it should be) will be transformed to
3=Log(10^3)-Logx, now tell me why is my method wrong

 

[Streetwolf]

Even so,

0 = -log x

x = 1

You can't take the log of a negative number.

And 0 * a negative number still = 0.

If you had 0 = log(-x) then you'd have x = -1.

 

[113zami]

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