A Logarithm that can't be solved

R

Relz

New member
Joined
Nov 20, 2011
Messages
49
I have a homework question asking for me to give a logarithmic equation that can't be solved. I would say log(-3) but it's not an equation. Would it work if I did this:

Log (5) + Log (0) = Log (5)(0) = Log 0

Log 0 cannot be solved because you can't take the log of zero.
 
JeffM said:
It is always possible that your text or problem sheet has a typographical error. Before assuming that you have inadvertently been given a senseless question, would you please give the problem EXACTLY as stated in your text.
Click to expand...

Sure!

"Give an example of a logarithmic equation that cannot be solved and explain why it cannot be solved. (3 marks)"
 
Since log(0) is not defined, I do not see your example as a valid equation.

It is like writing log(x) = your mother. There is obviously no solution because the equation does not make sense.


I think that they want a valid equation (containing at least one logarithmic term) that has no solutions.

One strategy would be to consider two logarithmic graphs that do not intersect.


For example, let's say that some exercise asks for an unsolvable exponential equation, instead.

I know that the graph of y = e^x lies entirely above the x-axis.

I know that the graph of y = -e^x lies entirely below the x-axis.

This is enough to know that e^x never equals -e^x; therefore, an exponential equation with no solutions is:

e^x = -e^x


Now, the graph of y = log(x) lies both above and below the x-axis, so log(x) = -log(x) won't work for this exercise.

However, the graph of y = log(x) does lie entirely to the right of the y-axis.

Does this fact (along with my example) give you any ideas?
 
mmm4444bot said:
Since log(0) is not defined, I do not see your example as a valid equation.

It is like writing log(x) = your mother. There is obviously no solution because the equation does not make sense.


I think that they want a valid equation (containing at least one logarithmic term) that has no solutions.

One strategy would be to consider two logarithmic graphs that do not intersect.


For example, let's say that some exercise asks for an unsolvable exponential equation, instead.

I know that the graph of y = e^x lies entirely above the x-axis.

I know that the graph of y = -e^x lies entirely below the x-axis.

This is enough to know that e^x never equals -e^x; therefore, an exponential equation with no solutions is:

e^x = -e^x


Now, the graph of y = log(x) lies both above and below the x-axis, so log(x) = -log(x) won't work for this exercise.

However, the graph of y = log(x) does lie entirely to the right of the y-axis.

Does this fact (along with my example) give you any ideas?
Click to expand...

Would that mean that -log(x) would lie on the left side of the y-axis?
 
Last edited:
OIC. Most schools don't wait until calculus to introduce logarithms; they introduce the material in a precalculus course.

Has your high school taught you about function transformations, yet?

In particular, I'm thinking about the effect that changing the sign of the function input has.

For example, what happens to the graph of f(x) = e^x, if we change the sign of the input to f(-x)?


If you have not yet learned about transformations, then here's a different approach.

Look at the graph of f(x) = log(x). There is lots of room to place a parabola (opening upward) with vertex in Quadrant II, such that the parabola and the log curve never meet.

If you can construct such a parabola (there are many), then your unsolvable equation would take this form:

ax^2 + bx + c = log(x)

What are you thinking now? :cool:
 
Relz said:
Would that mean that -log(x) would lie on the left side of the y-axis?
Click to expand...

You edited your post, while I was typing, heh, heh.

No, the graph of -log(x) does not lie to the left of the y-axis.

Change from log(x) to -log(x) flips the graph of log(x) across the x-axis.

Do you have graphing technology available?
 
mmm4444bot said:
You edited your post, while I was typing, heh, heh.

No, the graph of -log(x) does not lie to the left of the y-axis.

Change from log(x) to -log(x) flips the graph of log(x) across the x-axis.

Do you have graphing technology available?
Click to expand...

haha sorry! Based on your first post, I came up with log(x)= x^2 + 1
 
Last edited:
They want an equation (containing at least one logarithmic term) that is not true for any value of x.

In other words, the equation has no solution; it is false for all values of x.



Look at the graph of y = log(x).

Find another expression whose graph does not meet the curve of log(x) -- such as a parabola.

Set that expression equal to log(x), and you'll have a valid equation with no solutions.



Alternatively, what transformation do you need to apply to log(x) to flip the graph across the y-axis?

That transformed expression set equal to log(x) also provides an answer to this problem.
 
Relz said:
I came up with log(x) = x^2 + 1
Click to expand...

That works, for the equation part. Did you finish the explanation part?


Note: When you add information to your threads by editing prior posts, the thread is neither moved to the top of the index nor marked as containing unread material.

In other words, we won't be notified that you've edited your post. Hence, we may not see most of these edits. I noticed the change in your post above only because somebody else added a comment and I happened to glance upward while reading.

It's best to append new information to a thread by submitting a new post versus "hiding" it in prior posts. :cool:


PS: Please ignore my previous suggestion about transformations. I just re-read it, and I can't figure out what I was thinking! (Sorry if I wasted your time.)
 
mmm4444bot said:
That works, for the equation part. Did you finish the explanation part?


Note: When you add information to your threads by editing prior posts, the thread is neither moved to the top of the index nor marked as containing unread material.

In other words, we won't be notified that you've edited your post. Hence, we may not see most of these edits. I noticed the change in your post above only because somebody else added a comment and I happened to glance upward while reading.

It's best to append new information to a thread by submitting a new post versus "hiding" it in prior posts. :cool:


PS: Please ignore my previous suggestion about transformations. I just re-read it, and I can't figure out what I was thinking! (Sorry if I wasted your time.)
Click to expand...

okay, will do! As for the explanation part I said that it can't be solved because there is no value of "x" that will satisfy both sides of the equation.
 
Relz said:
As for the explanation part I said that it can't be solved because there is no value of "x" that will satisfy both sides of the equation.
Click to expand...

Hmmm. I don't think I would accept that as an explanation.

Stating that "no value of x satisfies both sides" is just repeating the statement "the equation has no solution" using different wording.

An explanation will state the reason why there is no solution.

Think about it. :cool:
 
You must log in or register to reply here.
Top