Making x the subject

How to make x the subject?

y = 0.1634Ln(x) + 0.0692

e.g. I want y to be 0.778659

*Sorry for forgetting to add the equation in the first place!!

Reply 1

The Bachelor

You're going to have to be much more specific...

What are you trying to do, exactly?

Reply 2

Catchetat

See above lol. Sorry :P

Reply 3

The Bachelor

So 0.778659 = 0.1634Ln(x) + 0.0692

How far do you get in solving for x?

Reply 4

Catchetat

Reply 5

ahmed.pilot

You should know that to make the x the subject, you must eliminate the logarithm. Because I'm not going to outright give you the answer, how can you go about eliminating the logarithm?

Reply 6

Catchetat

Right. I've never learnt about logarithms, that's the problem :P

Reply 7

fusionskd

Everything you need to learn [you might aswell know it anyway, incase you'll need it in future] Logarithms

Reply 8

Catchetat

Don't understand... but am really in the urge to know the answer to my question >.>

Reply 9

James

y=A\log x+B

\frac{y-B}{A}=\log x

Now apply the exponential function, which is the inverse of 'log':

\exp\left({\frac{y-B}{A}}\right)=x

Reply 10

HBKss

y = 0.1634ln(x) + 0.0692

yln(e) = ln[x(^0.1634)] + 0.0692ln(e)

ln(e^y) = ln[x(^0.1634)] + ln[e^(0.0692)]

ln(e^y) = ln[x(^0.1634)] + ln(1.071)

ln(e^y) = ln[1.071x(^0.1634)]

e^y = 1.071x(^0.1634)

(e^y)/1.071 = x(^0.1634)

now take power 1/0.1634 on both sides

final answer

[(e^y)/1.071](^6.12) =x

The procedure is this, there might be any calculational error in it