Proving equivalent functions with graphs

[skeptipus]

Hello

I have simplified this logarithmic function:

h(x)=0.25*log_2 ((8x-56)^(16)) - 12

to

g(x) = 4*log_2 (x-7)

These are equivalent functions. I need to prove that via graphing.

However, when I graph the original function (h(x)), I get something of a "V".
Whereas, g(x) gives me the characteristic graph of log functions.

Graphing h(x)=4*log_2 ((8x-56)) - 12 instead gives me the same graph as g(x) does.

My question is, what domain restrictions must I apply to h(x) for its graph to appear as g(x)'s?

Also, please kindly explain why does this happen? Are equivalent functions not truly equivalent?

Thank you very much for your time.

 

[Steven G]

Hello

I have simplified this logarithmic function:

h(x)=0.25*log_2 ((8x-56)^(16)) - 12

to

g(x) = 4*log_2 (x-7)

These are equivalent functions. I need to prove that via graphing.

However, when I graph the original function (h(x)), I get something of a "V".
Whereas, g(x) gives me the characteristic graph of log functions.

Graphing h(x)=4*log_2 ((8x-56)) - 12 instead gives me the same graph as g(x) does.

My question is, what domain restrictions must I apply to h(x) for its graph to appear as g(x)'s?

Also, please kindly explain why does this happen? Are equivalent functions not truly equivalent?

Thank you very much for your time.

h(x)=0.25*log_2 ((8x-56)^(16)) - 12 does equal h(x)=4*log_2 ((8x-56)) - 12
Where did you graph these functions? Maybe you typed it in wrong? Can you show us what you typed in exactly?
All your listed functions have the same domain.

 

[skeptipus]

It seems that you have two h(x) function--one being the original function.
I hope that it is clear that if h(x) and g(x) give the same graph, but the original h(x) gives a different graph, then h(x) and the original h(x) are NOT equal.

Can you give us the original function AND the exact statement of the program so we can figure out what is being asked????

I apologize for any ambiguity.

This is the complete question.

 

[Dr.Peterson]

Hello

I have simplified this logarithmic function:

h(x)=0.25*log_2 ((8x-56)^(16)) - 12

to

g(x) = 4*log_2 (x-7)

These are equivalent functions. I need to prove that via graphing.

However, when I graph the original function (h(x)), I get something of a "V".
Whereas, g(x) gives me the characteristic graph of log functions.

Graphing h(x)=4*log_2 ((8x-56)) - 12 instead gives me the same graph as g(x) does.

My question is, what domain restrictions must I apply to h(x) for its graph to appear as g(x)'s?

Also, please kindly explain why does this happen? Are equivalent functions not truly equivalent?

Thank you very much for your time.

The difference comes in moving the exponent outside of the log. The 16th power is always positive (except at x=7), so the domain of h is x≠7, while the domain of g is x>7. So the simplification did change the domain.

In order to keep the same domain, you should have written g(x) = 4*log_2 |x-7|. I often forget this!

 

[skeptipus]

The difference comes in moving the exponent outside of the log. The 16th power is always positive (except at x=7), so the domain of h is x≠7, while the domain of g is x>7. So the simplification did change the domain.

In order to keep the same domain, you should have written g(x) = 4*log_2 |x-7|. I often forget this!

Thank you very much, Dr. Peterson! :-D:-D

 

[skeptipus]

h(x)=0.25*log_2 ((8x-56)^(16)) - 12 does equal h(x)=4*log_2 ((8x-56)) - 12
Where did you graph these functions? Maybe you typed it in wrong? Can you show us what you typed in exactly?
All your listed functions have the same domain.

Thank you for your time, Jomo.
I was in the process of making a reply when Dr.Peterson had posted the answer to my question.

 

[Steven G]

In order to keep the same domain, you should have written g(x) = 4*log_2 |x-7|. I often forget this!

So do I!