# Thread: how to emdas in logatirthm?

1. ## how to emdas in logatirthm?

can somebody show me the proper step to solve for x

50/(1 + e^-x) = 4

first would be multiply denominator to 4,, .

correct answer is somewhere at -2.44

thanks

2. You can do this: $\displaystyle 50e^x = 4e^x + 4\; \Rightarrow \;46e^x = 4\; \Rightarrow \;e^x = \frac{2}{{23}}$

3. Originally Posted by aeroflix
can somebody show me the proper step to solve for x

50/(1 + e^-x) = 4

first would be multiply denominator to 4,, .

correct answer is somewhere at -2.44

thanks
Having first multiplied both sides by the denominator you have 50= 4(1+ e^{-x})= 4+ 4e^{-x} where I have used the "distributive law" to separate the term involving x from the constant term on the right. Now subtract 4 from both sides: 46= 4e^{-x}. Divide both sides by 4: 46/4= 23/2= e^{-x}. To get rid of the exponential now, do the "opposite"- take the natural logarithm of both sides: ln(23/2)= -x. Finally, multiply by -1.

Plato does it in a completely different way, first multiplying both numerator and denominator of the fraction on the left by e^x.