Question about exponential functions

[Tanish Shukla]

This question is from an exercise in an Edexcel Further Pure Mathematics book. Parts a, b, and c were fairly easy for me. For part a, I found the corresponding values of y using a calculator and in part b I drew the curve with no problems. In part (c) I rearranged the question:
[math]e^x=10[/math][math]1/2e^x=5[/math][math]1/2e^x+3=8[/math]Then used the graph to find the value of x for which the function was equal to 8.
However, I did not know where to start with part (d). I tried using the power rule of logs to bring 1-x to the power of 4, but I didn’t know what to do after that. Could someone explain how to solve part (d)?

 

[Steven G]

Did you solve for x??
(1+ln4)x=ln4

 

[Dr.Peterson]

This question is from an exercise in an Edexcel Further Pure Mathematics book. View attachment 38193Parts a, b, and c were fairly easy for me. For part a, I found the corresponding values of y using a calculator and in part b I drew the curve with no problems. In part (c) I rearranged the question:
[math]e^x=10[/math][math]1/2e^x=5[/math][math]1/2e^x+3=8[/math]Then used the graph to find the value of x for which the function was equal to 8.
However, I did not know where to start with part (d). I tried using the power rule of logs to bring 1-x to the power of 4, but I didn’t know what to do after that. Could someone explain how to solve part (d)?

First, the question is a little ambiguous, because of the lack of explicit indication of the argument of the log. I would interpret [imath]x=\ln4(1-x)[/imath] as [imath]x=\ln[4(1-x)][/imath], but one could also take it to mean [imath]x=[\ln(4)](1-x)[/imath], which I think is what @Steven G is doing. Does your class have a convention for the order of operations here?

Once you settle that, you want to rewrite this equation in exponential form, and then rearrange it so that one side is [imath]\frac{1}{2}e^x+3[/imath] and the other is the equation of a line. That will be the line you want to graph.

 

[Tanish Shukla]

Did you solve for x??
(1+ln4)x=ln4

I don't think the question is asking you to solve for x using algebra.
The question already asked to draw a curve of 1/2 e^x + 3 in parts a and b.
What it is asking is to rearrange it in such a way that you can draw a simple line on the graph and then figure out the solution to that question visually on the grid, if that makes sense.

 

[Tanish Shukla]

First, the question is a little ambiguous, because of the lack of explicit indication of the argument of the log. I would interpret [imath]x=\ln4(1-x)[/imath] as [imath]x=\ln[4(1-x)][/imath], but one could also take it to mean [imath]x=[\ln(4)](1-x)[/imath], which I think is what @Steven G is doing. Does your class have a convention for the order of operations here?

Once you settle that, you want to rewrite this equation in exponential form, and then rearrange it so that one side is [imath]\frac{1}{2}e^x+3[/imath] and the other is the equation of a line. That will be the line you want to graph.

I think it would be interpreted as [imath]x=[\ln(4)]*(1-x)[/imath]. I am pretty sure the book would have made it ln [4(1-x)] if it was a single argument, as I have seen the book doing this in other cases.

 

[Dr.Peterson]

I think it would be interpreted as [imath]x=[\ln(4)]*(1-x)[/imath]. I am pretty sure the book would have made it ln [4(1-x)] if it was a single argument, as I have seen the book doing this in other cases.

You might show us such an example to make sure; but we can trust your judgment.

Now do what I suggested, and try to do what you said:

What it is asking is to rearrange it in such a way that you can draw a simple line on the graph and then figure out the solution to that question visually on the grid, if that makes sense.

and as I said:

Once you settle that, you want to rewrite this equation in exponential form, and then rearrange it so that one side is [imath]\frac{1}{2}e^x+3[/imath] and the other is the equation of a line. That will be the line you want to graph.

Let's see your attempt. (I got a reasonable solution taking it my way, but don't see that it can be done the way you take it.)

 

[Otis]

don't see that it can be done [ using ln(4) ]

I tried, but ended up in a vicious circle, with either a ratio of exponentials on the left or an exponential factor in what's supposed to be the linear equation.

Exponentiating both sides of x = ln(4 – 4x) makes more sense (only two steps, after that). Perhaps the book is inconsistent with function notation.

 

[Tanish Shukla]

Exponentiating both sides of x = ln(4 – 4x)

I tried what Otis said to exponentiate both sides and it worked.
x = ln(4-4x)
Now raising to the power of e on both sides, we have:
e^x = 4-4x
Taking half of both sides:
1/2 e^x = 2-2x
Adding 3:
1/2 e^x + 3 = 5 - 2x
Now since I had the equation of the graph on the LHS I drew the line 5 - 2x on the same grid, and the curve and the line intersected to give me the right solution.
Thanks everybody!

 

[Dr.Peterson]

I tried what Otis said to exponentiate both sides and it worked.
x = ln(4-4x)
Now raising to the power of e on both sides, we have:
e^x = 4-4x
Taking half of both sides:
1/2 e^x = 2-2x
Adding 3:
1/2 e^x + 3 = 5 - 2x
Now since I had the equation of the graph on the LHS I drew the line 5 - 2x on the same grid, and the curve and the line intersected to give me the right solution.
Thanks everybody!

Yes, with this interpretation of the problem, it is easy.

I'd still be interested to see the evidence you mentioned that the book treats log a(b) as (log a)(b); I am interested in the variations of the order of operations, especially what people do without formal rules. (I discussed similar issues with trig functions here.)

 

[Tanish Shukla]

Yes, with this interpretation of the problem, it is easy.

I'd still be interested to see the evidence you mentioned that the book treats log a(b) as (log a)(b); I am interested in the variations of the order of operations, especially what people do without formal rules. (I discussed similar issues with trig functions here.)

The book covers logs quite briefly so there aren't many instances of your examples being used; however now that you mention this, I have realized that the use of log (ab) and (log a)(b) is used quite interchangeably in the book.
Here are some examples with and without parentheses, used in the book:


I suppose that given the "context" of the question, one would be able to differentiate (that was a horrible math pun) between log with and without the parentheses.
I also checked the answer key of the book and it turned out that my solution was indeed correct, implying that the book DID actually mean ln[4(1-x)].
I also skimmed the website that you linked briefly and found it quite interesting, almost as if people speak in different "dialects" of math, if you will.
If you are interested in the name of the book that this question is from, it is the Pearson Edexcel Further Pure Mathematics Student Book (ISBN 9780435044145).

 

[Dr.Peterson]

The book covers logs quite briefly so there aren't many instances of your examples being used; however now that you mention this, I have realized that the use of log (ab) and (log a)(b) is used quite interchangeably in the book.
Here are some examples with and without parentheses, used in the book:
View attachment 38222
View attachment 38223
I suppose that given the "context" of the question, one would be able to differentiate (that was a horrible math pun) between log with and without the parentheses.
I also checked the answer key of the book and it turned out that my solution was indeed correct, implying that the book DID actually mean ln[4(1-x)].
I also skimmed the website that you linked briefly and found it quite interesting, almost as if people speak in different "dialects" of math, if you will.
If you are interested in the name of the book that this question is from, it is the Pearson Edexcel Further Pure Mathematics Student Book (ISBN 9780435044145).

Your examples show that log xy is taken to mean log(xy), so it doesn't really support taking [imath]\log 4(1-x)[/imath] as anything other than [imath]\log(4(1-x))[/imath].

The (standard) order of operations rule they're following is that multiplication has precedence over logs; it's just easier to misread your problem than one with no parentheses at all.

Also, they clearly follow the tradition of putting parentheses around the argument of a log only when they feel like it, not for any consistent reason; they would be perfectly consistent to write [imath]\log_ax^3y^4z[/imath] instead of [imath]\log_a(x^3y^4z)[/imath]

The main lesson is, when you read a log not immediately followed by a parenthesis, consider the possibility that you may not be reading what was intended. And when you write a log, use parentheses unless it's clear no one can misread it!