Exponential equation

[cosmic]

I don't know what I'm doing wrong here any help would be greatly appreciated

I've included my working out for you to have a look at.

13^2x+10=10
ln 13^2x+10= ln 10
(2x+10) ln 13= ln 10
2x+10= ln 10/ln 13
2x=ln 10/ln 13 -10
x=1/2(ln 10/ln 13 -10)
x= -4.32 (3 sig. fig.)

Thanks in advance

 

[HallsofIvy]

I don't know what I'm doing wrong here any help would be greatly appreciated

I've included my working out for you to have a look at.

13^2x+10=10
ln 13^2x+10= ln 10
(2x+10) ln 13= ln 10
2x+10= ln 10/ln 13
2x=ln 10/ln 13 -10
x=1/2(ln 10/ln 13 -10)

Up to here you are correct.

x= -4.32 (3 sig. fig.)

But this is about 0.23 off! Check your calculator.

Thanks in advance

 

[cosmic]

Up to here you are correct.


But this is about 0.23 off! Check your calculator.

Thank you for your reply )

Sorry that should have said -4.55 (3 sig. fig.) But that's not what I meant when I entered the equation into a mathematical software I got the answer -2.05 (3 sig. fig.) so I'm wondering why that's the case.

 

[cosmic]

How can we answer that? We can't see what you entered...

https://www.google.ca/search?hl=en-......1ac.2j1.34.heirloom-hp..12.0.0.QhQuI5VqO-g

Sorry Denis, I can see why that might cause some confusion.

I've provided a link below. The answer it gives me is -2.04.... so one of us is wrong which is usually me so I thought I'd check with you guys

http://www.mathway.com/problem/MTY0NTk5OTY2

 

[Deleted member 4993]

I don't know what I'm doing wrong here any help would be greatly appreciated

I've included my working out for you to have a look at.

13^2x+10=10
ln 13^2x+10= ln 10
(2x+10) ln 13= ln 10
2x+10= ln 10/ln 13
2x=ln 10/ln 13 -10
x=1/2(ln 10/ln 13 -10)
x= -4.32 (3 sig. fig.)

Thanks in advance

just by observation 13 ≈ 10 → 2x + 10 ≈ 0 → x ≈ -5 → Your answer is "more accurate" than the "web-site answer"

 

[cosmic]

Well, all I get is the site...don't see anything you entered...

The link should have the question 13^2x+10=10 already entered you just have to click on answer to solve for x. If not you can manually enter the equation yourself and see what answer you get.

Thank you

 

[cosmic]

just by observation 13 ≈ 10 → 2x + 10 ≈ 0 → x ≈ -5 → Your answer is "more accurate" than the "web-site answer"

Thank you for your reply it just seems kind of strange to me that the two answers should be almost 2.5 apart.

 

[wjm11]

The answer it gives me is -2.04.... so one of us is wrong which is usually me so I thought I'd check with you guys

There is a simple solution; always plug your answer back into the original equation and see if it works.

In this case, your answer of -4.55 is correct. If you plug it in to the original expression, you get about 10.06, which is close to 10. If you plug the online answer in, it doesn't work.

 

[lookagain]

13^2x+10=10

ln 13^2x+10= ln 10

(2x+10) ln 13= ln 10

2x+10= ln 10/ln 13 $\displaystyle \ \ \ \ \$ <------- Here is where you started to need grouping symbols in particular. **

2x=ln 10/ln 13 -10

x=1/2(ln 10/ln 13 -10)

x= -4.32 (3 sig. fig.)

** cosmic, as usual for forum users, the error is due to the lack of grouping symbols.
The way it is shown by you here, and the way it must have been entered is that the
incorrect argument was 10/ln(13), instead of the two separate correct arguments
of ln(10)/ln(13) as a fraction.

Denis showed one of the correct forms of entering it in his link in post #4.

Look at this amendment:


13^2x+10 = 10

ln(13^2x+10) = ln(10)

(2x + 10)ln(13) = ln(10)

2x + 10 = [ln(10)]/[ln(13)]

2x = [ln(10)]/[ln(13)] - 10

x = (1/2){[ln(10)]/[ln(13)] - 10}

$\displaystyle x \ \approx \ -4.55$


- - - - - - - - - - - - - - - - - - - - - - - - - - - -


Here is an incorrect way of entering the problem and what you get when you enter the following
into a TI-30XIIS calculator or a TI-83 calculator, for example:

$\displaystyle (1/2)(ln(10/ln(13)) - 10) \ \approx \ -4.32$

 

[cosmic]

Got it this time...but had to hit "refresh"!
Strange indeed...suggest you send the site boss a note...

Yeah that's what I thought. Usually the answers are always correct so I'm not sure what seems to be the problem in this case.

Thank you

 

[cosmic]

There is a simple solution; always plug your answer back into the original equation and see if it works.

In this case, your answer of -4.55 is correct. If you plug it in to the original expression, you get about 10.06, which is close to 10. If you plug the online answer in, it doesn't work.

Thanks for that, I must admit I didn't think of this initially as I always seem to forget that I can check my answer in this manner.

All the best

 

[cosmic]

** cosmic, as usual for forum users, the error is due to the lack of grouping symbols.
The way it is shown by you here, and the way it must have been entered is that the
incorrect argument was 10/ln(13), instead of the two separate correct arguments
of ln(10)/ln(13) as a fraction.

Denis showed one of the correct forms of entering it in his link in post #4.

Look at this amendment:


13^2x+10 = 10

ln(13^2x+10) = ln(10)

(2x + 10)ln(13) = ln(10)

2x + 10 = [ln(10)]/[ln(13)]

2x = [ln(10)]/[ln(13)] - 10

x = (1/2){[ln(10)]/[ln(13)] - 10}

$\displaystyle x \ \approx \ -4.55$


- - - - - - - - - - - - - - - - - - - - - - - - - - - -


Here is an incorrect way of entering the problem and what you get when you enter the following
into a TI-30XIIS calculator or a TI-83 calculator, for example:

$\displaystyle (1/2)(ln(10/ln(13)) - 10) \ \approx \ -4.32$

Thanks for your reply. I can see that the grouping wasn't correct in the initial answer but I did get the correct answer in my first draft. I think it was a case of missing out a bracket when I entered the problem into my calculator a second time.

All the best.