Exponential Equations and Inequalities

[Lion]

Hi everyone,
I've finished the chapter of Exponential Equations and Inequalities of an algebra book . But the final 7 questions are difficult. it would be great if you could please tell me how to solve them. I tried more than one one hour to solve them but I got stocked...and I don't have a teacher.

1) (2^x)= (x^2)
2) (e^x)= ln(x) +5
3) (e^(x^0.5))=x+1
4) (e^-x)-x(e^-x)> or = 0
5) 3(x-1)< (2^x)
6) (e^x)<(x^3)-x
7) the inverse of : ((e^x)-(e^-x))/2

I've the final answers, but I have not figured out how to get them.

Thanks

 

[Deleted member 4993]

Hi everyone,
I've finished the chapter of Exponential Equations and Inequalities of an algebra book . But the final 7 questions are difficult. it would be great if you could please tell me how to solve them. I tried more than one one hour to solve them but I got stocked...and I don't have a teacher.

1) (2^x)= (x^2)
2) (e^x)= ln(x) +5
3) (e^(x^0.5))=x+1
4) (e^-x)-x(e^-x)> or = 0
5) 3(x-1)< (2^x)
6) (e^x)<(x^3)-x
7) the inverse of : ((e^x)-(e^-x))/2

I've the final answers, but I have not figured out how to get them.

Thanks

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33

 

[pka]

Hi everyone,
I've finished the chapter of Exponential Equations and Inequalities of an algebra book . But the final 7 questions are difficult. it would be great if you could please tell me how to solve them. I tried more than one one hour to solve them but I got stocked...and I don't have a teacher.
1) (2^x)= (x^2)
2) (e^x)= ln(x) +5
3) (e^(x^0.5))=x+1
4) (e^-x)-x(e^-x)> or = 0
5) 3(x-1)< (2^x)
6) (e^x)<(x^3)-x
7) the inverse of : ((e^x)-(e^-x))/2
I've the final answers, but I have not figured out how to get them.

I have no idea how you are expected to solve these.
For example in 1), \(\displaystyle x=2~ \&~x=4\) are both solutions by inspection. But there are other solutions. But we can find them only by very advanced methods (Lambert functions).

So you need to tell us what topics you have studied in this project.

 

[Harry_the_cat]

Hi everyone,
I've finished the chapter of Exponential Equations and Inequalities of an algebra book . But the final 7 questions are difficult. it would be great if you could please tell me how to solve them. I tried more than one one hour to solve them but I got stocked...and I don't have a teacher.

1) (2^x)= (x^2)
2) (e^x)= ln(x) +5
3) (e^(x^0.5))=x+1
4) (e^-x)-x(e^-x)> or = 0
5) 3(x-1)< (2^x)
6) (e^x)<(x^3)-x
7) the inverse of : ((e^x)-(e^-x))/2

I've the final answers, but I have not figured out how to get them.

Thanks

If you have a graphics calculator you can graph both sides, find the intersection points and interpret your answer.

They are difficult to do algebraically.

 

[Lion]

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33

Yeah sure thanks for letting me know about it, you're right...I did't know about the "Read before posting" as it its my first day here.
Here is my work:

1) x ln2 = (2 ln x)
and I'm stuck.

2) e^x = ln x +ln (e^5)
e^x = ln (x (e^5))
and I'm stuck.

3) ln (e^x) = ln ((x+a)^2)
2= x/(ln (x+1))
and I'm stuck.

4) let e^-1 =u
u-xu =0
u(1-x)=0
u=0 is undefined or x=1
solution (- infinity, 1)
I think this is correct?

5) (x-a) ln 3 = x ln 2
-ln 3 = x(ln 2 - ln 3)
x = (-ln x)/ (ln 2 - ln 3)
x = 2.709
solution: (-infinity, 2.709)
I think it is ok?

6) e^x = x(x-1)(x-)
0= ((x(x-1)(x-1))^(x-1) )- e
there are 3 or 1 positive zeros, non negative zeros
possible zeros:e
by dividing, it does not work, and I could not find the possible zeros
and I'm stuck.

7) interchanging x by y
x= ((e^y)- ((e^y)^-1))/2
put e^y = u
x=(u-(u^-1))/2
2x =((u-1)(u+1))/u
and I'm stuck.

Thanks

 

[Lion]

I have no idea how you are expected to solve these.
For example in 1), \(\displaystyle x=2~ \&~x=4\) are both solutions by inspection. But there are other solutions. But we can find them only by very advanced methods (Lambert functions).

So you need to tell us what topics you have studied in this project.

Hi...I don't know if the book would expect me to solve them by inspection.

Ok let me tell you, I am studying a book called College Algebra by Carl Stitz, alone to prepare for my next academic year (master in theoretical economics) because I have Algebra and calculus with other courses as prerequisites and I have not taken them before. I'm doing well, I have finished those 5 out of 9 chapters alone: relations and functions, Linear and quadratic functions, polynomials, rational functions, further topics in functions.

Now, I'm doing the sixth chapter exponential and log functions.

Thanks

 

[Lion]

If you have a graphics calculator you can graph both sides, find the intersection points and interpret your answer.

They are difficult to do algebraically.

Hello Harry, yes I have a graphics calculator...but the exercise is expecting me to solve the questions analytically without a calculator...I have the final answers at the end of the book, I'm interested in how to solve them algebraically.

thanks again

 

[Deleted member 4993]

Yeah sure thanks for letting me know about it, you're right...I did't know about the "Read before posting" as it its my first day here.
Here is my work:

1) x ln2 = (2 ln x)
and I'm stuck.

2) e^x = ln x +ln (e^5)
e^x = ln (x (e^5))
and I'm stuck.

3) ln (e^x) = ln ((x+a)^2)
2= x/(ln (x+1))
and I'm stuck.

4) let e^-1 =u
u-xu =0
u(1-x)=0
u=0 is undefined or x=1
solution (- infinity, 1)
I think this is correct?

5) (x-a) ln 3 = x ln 2
-ln 3 = x(ln 2 - ln 3)
x = (-ln x)/ (ln 2 - ln 3)
x = 2.709
solution: (-infinity, 2.709)
I think it is ok?

6) e^x = x(x-1)(x-)
0= ((x(x-1)(x-1))^(x-1) )- e
there are 3 or 1 positive zeros, non negative zeros
possible zeros:e
by dividing, it does not work, and I could not find the possible zeros
and I'm stuck.

7) interchanging x by y
x= ((e^y)- ((e^y)^-1))/2
put e^y = u
x=(u-(u^-1))/2
2x =((u-1)(u+1))/u
and I'm stuck.

Thanks

All these are non-linear equations of infinite order. So these do not have "closed-form" algebraic solutions. You will need to use numerical method/s to get approximate solutions.

The 7) is almost done:

x = (u² - 1)/(2u)

u² - 2ux - 1 = 0 ...... Quadratic equation

u = 1 ± √(x²+1)

e^y = 1 ± √(x²+1)

y = ln[1 ± √(x²+1)] → y = ln[1 + √(x²+1)] .... because (x² + 1)>1

 

[pka]

Ok let me tell you, I am studying a book called College Algebra by Carl Stitz, alone to prepare for my next academic year (master in theoretical economics) because I have Algebra and calculus with other courses as prerequisites and I have not taken them before. I'm doing well, I have finished those 5 out of 9 chapters alone: relations and functions, Linear and quadratic functions, polynomials, rational functions, further topics in functions.
Now, I'm doing the sixth chapter exponential and log functions.

I do not know that textbook nor the author (there are almost 500+ college algebra texts on the market).

Here is the analytic solution for #1. I would be in total disbelief if any basic algebra text contained material on Lambert's W-functions.

 

[Lion]

thanks Khan, this is very clear...
for 1-6, I understood that, it is not possible to solve them analytically...correct? I wonder if would you give me to which book, or methods should I use to solve them?

 

[Lion]

Thanks Pka...the book is online and it is quiet good. So you think that those problems 1-6 are very advanced? Should I just ignore them and carry on to next chapters? as I have to do finish another courses of Linear algebra, beginner, intermediate and advanced calculus...alongside with more ones before starting the next course.

Thanks again

 

[Deleted member 4993]

thanks Khan, this is very clear...
for 1-6, I understood that, it is not possible to solve them analytically...correct? I wonder if would you give me to which book, or methods should I use to solve them?

One methods that can be used is known as Newton-Raphson iterative method.

 

[Ishuda]

thanks Khan, this is very clear...
for 1-6, I understood that, it is not possible to solve them analytically...correct? I wonder if would you give me to which book, or methods should I use to solve them?

The way I am reading the questions, (4) and (7) are solvable algebraically:
(4) \(\displaystyle e^{-x}\, -\, x\, e^{-x}\, \ge\, 0\)
Write as
\(\displaystyle e^{-x}\, [1\, -\, x\, ] \ge\, 0\)
and since e^-x is always positive, we have \(\displaystyle -\infty\, \lt\, x\, \le\, 1\) which differs from your solution only in that x=1 is included as a solution [an interval closed on the right] wheras your interval is open on the right.
(7)Get the inverse of y = \(\displaystyle \frac{e^x\, -\, e^{-x}}{2}\)
Interchanging x and y and letting t = e^y we can write the inverse as
x = \(\displaystyle \frac{t\, -\, t^{-1}}{2}\)
or
\(\displaystyle 2\, x\, t\, =\, t^2\, -\, 1\)
This is just a quadratic equation in t. So solve the quadratic equation for t to get
t = \(\displaystyle x\, \pm\, \sqrt{x^{2}\, +\, 1}\)
We note that, along the way to the explanation of why we have two different solutions, that
\(\displaystyle x\, -\, \sqrt{x^{2}\, +\, 1}\) = -[\(\displaystyle \frac{1}{x\, +\, \sqrt{x^{2}\, +\, 1}}\)]

Oh, and remember that t=e^y, so take a log

 

[Lion]

The way I am reading the questions, (4) and (7) are solvable algebraically:
(4) \(\displaystyle e^{-x}\, -\, x\, e^{-x}\, \ge\, 0\)
Write as
\(\displaystyle e^{-x}\, [1\, -\, x\, ] \ge\, 0\)
and since e^-x is always positive, we have \(\displaystyle -\infty\, \lt\, x\, \le\, 1\) which differs from your solution only in that x=1 is included as a solution [an interval closed on the right] wheras your interval is open on the right.
(7)Get the inverse of y = \(\displaystyle \frac{e^x\, -\, e^{-x}}{2}\)
Interchanging x and y and letting t = e^y we can write the inverse as
x = \(\displaystyle \frac{t\, -\, t^{-1}}{2}\)
or
\(\displaystyle 2\, x\, t\, =\, t^2\, -\, 1\)
This is just a quadratic equation in t. So solve the quadratic equation for t to get
t = \(\displaystyle x\, \pm\, \sqrt{x^{2}\, +\, 1}\)
We note that, along the way to the explanation of why we have two different solutions, that
\(\displaystyle x\, -\, \sqrt{x^{2}\, +\, 1}\) = -[\(\displaystyle \frac{1}{x\, +\, \sqrt{x^{2}\, +\, 1}}\)]

Oh, and remember that t=e^y, so take a log

Thanks man, this helps me...the only part that I have not understood is how:

x−x2+1‾‾‾‾‾‾‾√ = -[1x+x2+1√]

I don't figure out why they are equal?
would you tell me what do you use to write a proper equation, which software? (I'v an linux ubuntu)
Thanks a again:cool:

 

[Deleted member 4993]

Thanks man, this helps me...the only part that I have not understood is how:

x−x2+1‾‾‾‾‾‾‾√ = -[1x+x2+1√]

I don't figure out why they are equal?
would you tell me what do you use to write a proper equation, which software? (I'v an linux ubuntu)
Thanks a again:cool:

[x + √(x² + 1)][x - √(x² + 1)] = x² - [√(x² + 1)]² = -1

Please work it out with pencil and paper (and eraser) - not just stare at the screen!!

 

[Ishuda]

Thanks man, this helps me...the only part that I have not understood is how:

x−x2+1‾‾‾‾‾‾‾√ = -[1x+x2+1√]

I don't figure out why they are equal?
would you tell me what do you use to write a proper equation, which software? (I'v an linux ubuntu)
Thanks a again:cool:

Linux Mint here and it doesn't make a difference. There are many places documenting the Tex/LaTex 'language' around the web but possibly the best way to learn is just to look at the examples here and find a site you are comfortable with to extend your knowledge. Do a 'Reply To' to see examples [and then close the page if you don't want to post]. The major difference between here and some other sites are the tags used to wrap the LaTex code. The tag pair used here is [t e x] and [/t e x] without the spaces. For example, the square root of 1 plus x squared is written as \sqrt{1+x^2} but you must wrap the code in the tag pair used here, i.e

[tex]\sqrt{1+x^2}[/tex]

results in \(\displaystyle \sqrt{1+x^2}\)