Domain of natural log

alyren

Junior Member
Joined
Sep 9, 2010
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find the domain of the function

f(x) = ln(1/x-10)

i set the equation to f(x)= Ln1-Lnx-10

then i dont know, same as below

f(x) = log base10 [(x+2)/(x-3)]
f(x)= log base 10^x+2 - log base 10^x-3
 
alyren said:
find the domain of the function

f(x) = ln(1/x-10)

i set the equation to f(x)= Ln1-Lnx-10

then i dont know, same as below

f(x) = log base10 [(x+2)/(x-3)]
f(x)= log base 10^x+2 - log base 10^x-3

First tell us what is the domain of:

f(x) = ln(x)
 
alyren said:
would that be all real numbers,except 10?
what is the domain of:

f(x) = ln(x)

It would be x > 0 [meaning all the real numbers greater than 0]

So in your case {f(x) = ln [1/(x-10)]} the domain is (x - 10) > 0
 
what about the (1/x-10),
x can't equal to 10?

can you show me step to solve it?
 
alyren said:
f(x) = ln(1/x-10)

Your typing means this:

\(\displaystyle f(x) \;=\; ln \left ( \frac{1}{x} \;-\; 10 \right )\)

Is that what you intend ?

i set the equation to f(x)= Ln1-Lnx-10

ln(1) - ln(x) - 10 does not equal ln(1/x - 10)

Maybe function f is defined this way:

\(\displaystyle f(x) \;=\; ln \left ( \frac{1}{x - 10} \right )\)

If so, then we show the denominator by typing grouping symbols.

f(x) = ln(1/[x - 10])

Then you can write:

f(x) = ln(1) - ln(x - 10)

This can be simplified because ln(1) is a constant.

Do you know? Subtracting 10 from x causes the graph to shift 10 units to the right.

If you're familiar with the graph and domain of ln(x), then the graph and domain of ln(x - 10) should be clear to you.

If you're not familiar with the graph and domain of ln(x), then perhaps you are not ready for this exercise.

 
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