A little help (maths - again)

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Just what i want to be doing on a bank holiday!

So i have said formula:

E=0.0615log10[4]/[K]

I am having trouble making K the subject as i don't no much about logs in this context. Can anyone give me any hints?

Much appreciated,

Luke.
 
If its E = (0.0615log10[4]) / [K]

then surely it's just K = (0.0615log10[4]) / [E]

however if it is. E = 0.0615log10(4/k)

then....
E = 0.0615log10(4) - 0.0615log10(k)

0.0615log10(k) = 0.0615log10(4) - E

log10(k) = log10(4) - (E/0.0615)

k = 10^(log10(4) - (E/0.0615)

k = 10^(log10(4)) / 10^(E/0.0615)

k = 4 / (E/0.0615)

k = (4*0.0615) / E

k = 0.246 / E


Is this correct?
 
Last edited:
E = 0.0615log10(4/k).

let a = 0.0615.

E/a = log_10(4/k)

10^(E/a) = 4/k

k = 4 / 10^(E/a) = 4.10^(-E/a)

(. = multiply)


so Mason, no I think you went wrong when you got rid of '10^' in 10^(E/0.0615)
 
Last edited:
he wont need to know anything about logs in one context so it should be obvious...

op: do you know what the inverse operation of log_10(x) is?

No, i think that maybe the problem.. Could you briefly explain?

First of all i did what mason did in his first re-arragment but i dont think its that simple! E is 0.1 btw i forgot to include it..
 
10^(log(x)) = x
log(10^x) = x

inverse of a function, f, call it f` => f( f'(x) ) == x.

therefore if you want to 'pull x out' of y = log( 13/x ), you would want to apply the inverse of log.:

10^y = 10^(log(13/x))

and then use the fact that 10^log(x) = x to get,
10^y = 13/x
...
 
E = 0.0615log10(4/k).

let a = 0.0615.

E/a = log_10(4/k)

10^(E/a) = 4/k

k = 4 / 10^(E/a) = 4.10^(-E/a)

(. = multiply)


so Mason, no I think you went wrong when you got rid of '10^' in 10^(E/0.0615)

You are correct, I ignore the 10^ on the denominator d'oh!
 
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