Question

Logarithm and Functions Problem:

Answer 1

If f(x) =\[\log_{a} x \] and
\[h \neq0\],
show that :
\[\frac{ f(x+h)-f(x) }{ h }=\log_{a} (1+\frac{ h }{ x })^{\frac{ 1 }{ h }}\]

Answer 2

As a start, take the left hand side of you need to show and plug in the given function

Answer 3

thats the part I'm having trouble with

Answer 4

do i just add h to log_a * x?

Answer 5

for the first one? f(x+h)

Answer 6

\(\log_a \) is function notation like \(\sin \), it neesd an argument to make sense :

\(\log_a (x)\) is the logarithm of \(x\) relative to base \(a\)

Answer 7

so what would be the result of f(x+h) after you plug in f(x)?

Answer 8

simply replace \(x\) by \(x+h\) :

\(f(x) = \log_a(x)\)
\(f(x+h) = \log_a(x+h)\)

Answer 9

ok I've plugged in everything now how do i prove it

Answer 10

\[\begin{align}
\frac{ f(x+h)-f(x) }{ h } &= \dfrac{\log_a(x+h)-\log_a(x)}{h}\\~\\
\end{align}\]

Answer 11

what properties do you know about logarithms ?

Answer 12

most of them: Change of base, the sum and difference etc.

Answer 13

very good :) time to use them..
use difference of logarithms property on numerator for now

Answer 14

ok then what

Answer 15

step by step, first do that and the next step might become obvious :)

Answer 16

oh ok i got it

Answer 17

thanks!

Answer 18

\[\begin{align}
\frac{ f(x+h)-f(x) }{ h } &= \dfrac{\log_a(x+h)-\log_a(x)}{h}\\~\\
&=\dfrac{\log_a\left(\dfrac{x+h}{x}\right)}{h}\\~\\
&=\dfrac{\log_a\left(1+\dfrac{h}{x}\right)}{h}\\~\\
&=\log_a\left(1+\dfrac{h}{x}\right)^{\frac{1}{h}}\\~\\

\end{align}\]

Answer 19

Can you help me on this one too :
Find the domain of \[f(x)=\log_7(\frac{ x+1 }{ x })\]

Answer 20

logarithm can only digest positive numbers

Answer 21

so the stuff inside parenthesis must always stay positive :

\[\dfrac{x+1}{x} \gt 0 \]

solve \(x\)

Answer 22

http://www.wolframalpha.com/input/?i=solve+%28x%2B1%29%2Fx+%3E+0

Answer 23

Oh, nevermind..., my question to you: I mentally thought of it as:\[\frac{1}{h}\cdot \log_a\left(1+\frac{h}{x}\right)\]which was the "k" you just mentioned, haha. Sorry for the redundancy!

Answer 24

Ahh yes yes thats a pretty interesting property if we think about why it has to be true

Answer 25

This loosely reminds me of the exponential growth function.