SOLUTION: Hello. I have the following question to answer but I have not covered anything like this before and do not know where to start... "find a relationship for W from the current flow e

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Hello. I have the following question to answer but I have not covered anything like this before and do not know where to start... "find a relationship for W from the current flow e      Log On


   

Question 718791: Hello. I have the following question to answer but I have not covered anything like this before and do not know where to start... "find a relationship for W from the current flow equation +i=%28kT%5E2%29+e%5E%28%28W-E%29%2FkT%29+ ". I honestly do not know where on earth to start. Any help in kick starting me off in the right direction or an explanation into what is required will be greatfully appreciated! I thank you in advance
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
I believe that "find a relationship for W" means the same as "solve for W". And "solve for W" means to transform the equation so that W is by itself on one side. So our task is to use algebra to transform the given equation so that it is in the form:
W = some-expression
or
some-expression = W

In the given equation:
+i=%28kT%5E2%29+e%5E%28%28W-E%29%2FkT%29+
the W is in the middle of a big mess on the right side. It will take a while to get W by itself. First we will isolate the base, e, and its exponent by dividing both sides by kT%5E2%29:
+i%2F%28kT%5E2%29=+e%5E%28%28W-E%29%2FkT%29+
The next step is the hardest. We need to get W out of the exponent. To do this we use logarithms. While logarithms of any base may be used, we will get a simpler expression if we match the logarithm's base to the base of the exponent. So we will use base e logs, "ln":
+ln%28i%2F%28kT%5E2%29%29=+ln%28e%5E%28%28W-E%29%2FkT%29%29+
Next we use a property of logarithms, log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29, which allows us to move the exponent of the argument out in front. (It is this property that is the very reason we use logs in situations like this. It allows us to move the exponent, which has the variable we are solving for, out in front where we can "get at it" using "regular" algebra.) Using this property on the log on the right side we get:
+ln%28i%2F%28kT%5E2%29%29=+%28%28W-E%29%2FkT%29%2Aln%28e%29+
Since ln(e) = 1 this becomes:
+ln%28i%2F%28kT%5E2%29%29=+%28%28W-E%29%2FkT%29+
The rest is simple. Multiply by kT:
+kT%2Aln%28i%2F%28kT%5E2%29%29=+W-E+
And add E:
+kT%2Aln%28i%2F%28kT%5E2%29%29+%2B+E+=+W+