SOLUTION: use the laws of logarithms to show that for f(x)=log(sub b) x f(x+h) - f(x)/ h = 1/h log(sub b) (1+h/x) = 1/x log(sub b) (1+h/x)^x/h

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: use the laws of logarithms to show that for f(x)=log(sub b) x f(x+h) - f(x)/ h = 1/h log(sub b) (1+h/x) = 1/x log(sub b) (1+h/x)^x/h      Log On


   

Question 613380: use the laws of logarithms to show that for f(x)=log(sub b) x
f(x+h) - f(x)/ h = 1/h log(sub b) (1+h/x) = 1/x log(sub b) (1+h/x)^x/h
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29+=+log%28b%2C+%28x%29%29
f%28x%2Bh%29+=+log%28b%2C+%28x%2Bh%29%29

Now we want to find a way to transform the right side into:
%281%2Fh%29log%28b%2C+%281+%2B+h%2Fx%29%29%29
and then into:
%281%2Fx%29log%28b%2C+%28%281+%2B+h%2Fx%29%5E%28x%2Fh%29%29%29
I mention this because it can be very helpful to keep your eye on where you want to end up. It can help direct your thinking toward actions that help you reach your goal.

For example, seeing that we want a factor of 1/h and seeing that you have a division by h, you might easily recognize that since division and multiplying by a reciprocal are the same we can immediately rewrite

as

Each of the following steps are based on my thinking of a way to take part of what I have and figuring out a way to turn it into part of what I want.

I have two logs but I only want one. SO I use the log%28a%2C+%28p%29%29+-+log%28b%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29 property to combine the two logs I have into one:
%28f%28x%2Bh%29-f%28x%29%29%2Fh+=+%281%2Fh%29%28log%28b%2C+%28%28x%2Bh%29%2Fx%29%29%29

I have one term, a fraction, in the argument of the log. I want two. "Un-adding" the fraction gives us:
%28f%28x%2Bh%29-f%28x%29%29%2Fh+=+%281%2Fh%29%28log%28b%2C+%28x%2Fx%2Bh%2Fx%29%29%29
which simplifies to:
%28f%28x%2Bh%29-f%28x%29%29%2Fh+=+%281%2Fh%29%28log%28b%2C+%281%2Bh%2Fx%29%29%29
which is the intermediate goal!

Now on to out final goal. The final expression does not have a (1/h) in front. Using another property of logarithms, q%2Alog%28a%2C+%28p%29%29+=+log%28a%2C+%28p%5Eq%29%29, to move the coefficient into the argument as its exponent:
%28f%28x%2Bh%29-f%28x%29%29%2Fh+=+log%28b%2C+%28%281%2Bh%2Fx%29%5E%281%2Fh%29%29%29
The next part is the trickiest. We have an exponent of 1/h in the argument of the log. We want an argument of x/h. We want, in effect, to be able to multiply the exponent by x. But any operation that just multiplies the exponent by x would change the value of the expression. So we need to find a way to introduce a factor of x that will not change the value of the expression. Here's the logic of how:
  • We need, in effect, to multiply the exponent. The only time exponents get multiplied is when you raise a power to a power. So we will be raising the argument of the log, which already has an exponent, to some power.
  • The only power of something that does not change its value is 1. So we will be raising the argument of the log to a power that is equal to a 1
  • We want this power that is equal to 1 to have a factor of x in it. The simplest expression that has a factor of x and is equal to 1 is: (x)*(1/x)
  • Not only does the (x)*(1/x) give us the factor of x we want, but it also brings in a (1/x) that we would like to see end up in front of the log!
So our next step is to raise the argument of the log to the (x)*(1/x) power:

Now we use our exponent rule for raising a power to a power in ways that move us toward our goal. First we separate the (x) and (1/x) factors:

so that we can then combine just the x and 1/h exponents:


And finally, we have an exponent of 1/x in the argument and we want a 1/x in front. Using the q%2Alog%28a%2C+%28p%29%29+=+log%28a%2C+%28p%5Eq%29%29 property of logarithms (this time from right to left) we can simply move the 1/x from where we don't want it to where we do want it:

And we're finished!