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# Maths: Logs watch

1. Considering
f(x)=2log(2-x)
and
g(x)=log(2-x)^2

What is the domain for each function?
Is the domain the same for each function?
2. Maybe look at (not find because you might unable to get a valid value) f(3) and g(3) will help you understand the situation better. I think g(x) is equal to log((x-2)^2) right?

3. (Original post by begbie68)
Considering
f(x)=2log(2-x)
and
g(x)=log(2-x)^2

What is the domain for each function?
Is the domain the same for each function?
What base? You haven't even told us what base it is in.
What is the value of log(-2), log (-3)?
Can you take logs of a negative number?

Using the power rule of f(x), what do you get?
4. (Original post by _JB_)
Maybe look at (not find because you might unable to get a valid value) f(3) and g(3) will help you understand the situation better. I think g(x) is equal to log((x-2)^2) right?

Thanks for prompt response, JB
Yes, you're correct about g(x)... I don't think your second set of brackets are required, since (log(2-x))^2 would be bracketed thus.

I know how to find f(3) and g(3), and they are given as equivalents/identities in
alog(b) = log(b)^a

I'm more interested in your response to the final part of my question.... 'evidence/justification'
5. (Original post by Chittesh14)
What base? You haven't even told us what base it is in.
What is the value of log(-2), log (-3)?
Can you take logs of a negative number?

Using the power rule of f(x), what do you get?

Using the power rule of f(x), what do you get?[/QUOTE]

Hi, Chittesh, base is irrelevant to the domain, but if one is not given, then by default it is base 10, or ANY base. That's why I didn't give one, but thanks for your comment on that.

and yes, you're correct in thinking that f(x) is identical to g(x) by
alog(b) = log(b)^a

Most interested in evidence/justification if you can ....
6. (Original post by begbie68)
Hi, Chittesh, base is irrelevant to the domain, but if one is not given, then by default it is base 10, or ANY base. That's why I didn't give one, but thanks for your comment on that.

and yes, you're correct in thinking that f(x) is identical to g(x) by
alog(b) = log(b)^a

Most interested in evidence/justification if you can ....
Fair enough, the shape would be different according to the base lol. No worries, I like your understanding 😀.

Yeah, what is the domain of f(x)? What is the domain of g(x)?
Reason - because they're identical functions lol.
7. (Original post by Chittesh14)
Fair enough, the shape would be different according to the base lol. No worries, I like your understanding 😀.

Yeah, what is the domain of f(x)? What is the domain of g(x)?
Reason - because they're identical functions lol.
"shape" is actually the same (for logs of different (positive/real) bases). but no matter.

but the domains as separate functions are actually different... look more closely
8. (Original post by begbie68)
"shape" is actually the same. but no matter.

but the domains as separate functions are actually different... look more closely
Sorry, I didn't look properly. I meant the shape is different if the base is different and not taken as 10 lol.
Yes, what you have to realise is again the rule that you can't take log of a negative number
g(x) is the same as f(x) as a function.
If you draw it out - it is combining both f(x) and it's reflection in x = 2.

It has an asymptote at x = 2 because you can't take log of 0.
Remember, log(2-x)^2 means it will always have a positive result regardless for all real x values, except 0 where it gives log(0) and its undefined.
9. (Original post by begbie68)
Thanks for prompt response, JB
Yes, you're correct about g(x)... I don't think your second set of brackets are required, since (log(2-x))^2 would be bracketed thus.

I know how to find f(3) and g(3), and they are given as equivalents/identities in
alog(b) = log(b)^a

I'm more interested in your response to the final part of my question.... 'evidence/justification'
alog(b) = log(b)^a only applies if b is positive. f(x) is not identical to g(x). That is why i want you to investigate f(3) and g(3).
Where f(3) = 2log(2-3) = 2log(-1) ??? How about g(3).
10. (Original post by Chittesh14)
Sorry, I didn't look properly. I meant the shape is different if the base is different and not taken as 10 lol.
Yes, what you have to realise is again the rule that you can't take log of a negative number
g(x) is the same as f(x) as a function.
If you draw it out - it is combining both f(x) and it's reflection in x = 2.

It has an asymptote at x = 2 because you can't take log of 0.
Remember, log(2-x)^2 means it will always have a positive result regardless for all real x values, except 0 where it gives log(0) and its undefined.
Again, Chittesh, 'shapes' are the same for different (positive/real) bases.
But, no matter.

I know what the graphs of f(x) and g(x) look like, and have plotted them manually, and on various graphing tools/calculators.
The question is: why are the graphs DIFFERENT, and hence the domains are different, when the functions are defined as identical?

That's the final part of my question.... 'give reason/evidence to support'.

Thanks
11. (Original post by begbie68)
Nice one, JB. But where is it stated "ONLY IF b is positive"?
I am a bit confused by this sentence.
12. (Original post by begbie68)
Again, Chittesh, 'shapes' are the same for different (positive/real) bases.
But, no matter.

I know what the graphs of f(x) and g(x) look like, and have plotted them manually, and on various graphing tools/calculators.
The question is: why are the graphs DIFFERENT, and hence the domains are different, when the functions are defined as identical?

That's the final part of my question.... 'give reason/evidence to support'.

Thanks
Why are you trying to prove me wrong if I'm trying to help you ? lol

I am talking about the shape of the individual curves, not their comparison.
If f(x) was 2log(2-x) in base 10, it won't have the same shape as 2log(2-x) in base 5.

I just explained to you, it's because one can take x values of > 2 because the result gives log of a positive number.
For the other, if x > 2, log(2-x) is negative, so it is undefined for the function.

The reason why they're not identical is because we haven't got a domain for the curves, which is what we're trying to find.
f(x) would be identical to g(x) if x <= 2.
13. If I give you a similar question:

f(x) =
g(x) =

How would you do it? Think about it.
14. (Original post by _JB_)
I am a bit confused by this sentence.
Ok. Apologies.

So .... if this was part of a question along the lines of

2log(2-x) - log(ax+b) = c

which you solved and arrived correctly at x = x1, or x = x2

where one of x1,x2 is less than 2, is it still a valid root?
15. X > 2
16. (Original post by _JB_)
I am a bit confused by this sentence.
The identity works both ways...

so, if 2log2 = log4 = log 2^2 = log(-2)^2 = 2log(-2)

or not?
17. (Original post by begbie68)
Ok. Apologies.

So .... if this was part of a question along the lines of

2log(2-x) - log(ax+b) = c

which you solved and arrived correctly at x = x1, or x = x2

where one of x1,x2 is less than 2, is it still a valid root?
I need more context. Like what is a, b and c. Because (ax+b) and (x-2) cannot be negative as well.
18. (Original post by Chittesh14)
If I give you a similar question:

f(x) =
g(x) =

How would you do it? Think about it.
A couple of points to note here.
the graph of f(x) occurs only in the positive-positive (top right) quadrant if you want the bijective function. otherwise, both of the positive x quadrants.

g(x) could be graphed in all 4 quadrants and would have 2 lines of symmetry and rotational symmetry about origin, unless your definitions are more strictly defined, initially.
19. (Original post by begbie68)
A couple of points to note here.
the graph of f(x) occurs only in the positive-positive (top right) quadrant if you want the bijective function. otherwise, both of the positive x quadrants.

g(x) could be graphed in all 4 quadrants and would have 2 lines of symmetry and rotational symmetry about origin, unless your definitions are more strictly defined, initially.
g(x) could be graphed only in two quadrants, where y is positive.
20. (Original post by _JB_)
I need more context. Like what is a, b and c. Because (ax+b) and (x-2) cannot be negative as well.
sorry, should have said ... "either of x1,x2 is greater than 2"

understand extra context.

I'll try and offer a better example to more closely identify the intended issue.

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