2 variables system with natural logarithm

[Sap67]

Hey !
So I have the following system :

[MATH] x - 2y = 1 [/MATH][MATH] ln(x) + ln(y) = 3ln(5) - ln(10) + ln(4) [/MATH]
So I'll explain what I tried first :
First attempt :
Figured out that x and y can't be negatives if I remember the lessons correctly.
Then, I got rid of the ln function and got this :
[MATH] x + y = 5^3 - 10 + 4 [/MATH]
With the new system, I got the following answer :
[MATH] y = 118/3 [/MATH][MATH] x = 239/3 [/MATH]But that's wrong.

On my second attempt I tried this :
[MATH] ln(xy) = ln(5^3/10)(4) [/MATH]followed by
[MATH] xy = 50 [/MATH]
With the new system, I got the following answer :
[MATH] y = 49/3 [/MATH]But I didn't even bother to continue this as those are the approximate forms I should be getting (They're on the screenshot I sent ) :

Can someone tell me what I'm doing wrong ? I wish I could focus a bit more on this exercice and figure it out myself but I got a lot to study for tomorrow and I can't stay stuck on this too long.

 

[Dr.Peterson]

[MATH] x - 2y = 1 [/MATH][MATH] ln(x) + ln(y) = 3ln(5) - ln(10) + ln(4) [/MATH]
So I'll explain what I tried first :
First attempt :
Figured out that x and y can't be negatives if I remember the lessons correctly.
Then, I got rid of the ln function and got this :
[MATH] x + y = 5^3 - 10 + 4 [/MATH]

That is invalid. You can't just drop the logs.

Instead, condense the expression on each side to an expression consisting of just ln(...); then you can eliminate the lns by applying the inverse function to both sides ... as you do in your second attempt:

On my second attempt I tried this :
[MATH] ln(xy) = ln(5^3/10)(4) [/MATH]followed by
[MATH] xy = 50 [/MATH]
With the new system, I got the following answer :
[MATH] y = 49/3 [/MATH]But I didn't even bother to continue this as those are the approximate forms I should be getting (They're on the screenshot I sent ) :

Can someone tell me what I'm doing wrong ? I wish I could focus a bit more on this exercice and figure it out myself but I got a lot to study for tomorrow and I can't stay stuck on this too long.

Please show your work in solving this system. You should have had a quadratic equation to solve, and should not get a rational number.

 

[Deleted member 4993]

Hey !
So I have the following system :

[MATH] x - 2y = 1 [/MATH][MATH] ln(x) + ln(y) = 3ln(5) - ln(10) + ln(4) [/MATH]
So I'll explain what I tried first :
First attempt :
Figured out that x and y can't be negatives if I remember the lessons correctly.
Then, I got rid of the ln function and got this :
[MATH] x + y = 5^3 - 10 + 4 [/MATH]
With the new system, I got the following answer :
[MATH] y = 118/3 [/MATH][MATH] x = 239/3 [/MATH]But that's wrong.

On my second attempt I tried this :
[MATH] ln(xy) = ln(5^3/10)(4) [/MATH]followed by
[MATH] xy = 50 [/MATH]
With the new system, I got the following answer :
[MATH] y = 49/3 [/MATH]But I didn't even bother to continue this as those are the approximate forms I should be getting (They're on the screenshot I sent ) :

Can someone tell me what I'm doing wrong ? I wish I could focus a bit more on this exercice and figure it out myself but I got a lot to study for tomorrow and I can't stay stuck on this too long.

Then, I got rid of the ln function and got this :

You CANNOT do that!

On my second attempt I tried this :
ln(xy) = ln(5^3/10)(4)

You CANNOT do that!

This one should be:

ln(xy) = ln(\(\displaystyle \frac{5^3*4}{10})\)

Now you can get rid of 'ln' from both sides and get:

xy = 50

Continue.....

 

[Sap67]

You CANNOT do that!

You CANNOT do that!

This one should be:

ln(xy) = ln(\(\displaystyle \frac{5^3*4}{10})\)

Now you can get rid of 'ln' from both sides and get:

xy = 50

Continue.....

 

[Sap67]

For some reason I can't reply in the same message, so I'm just gonna have to double post. Sorry.
I pretty much get stuck here as I'm getting a negative radical.
I'm guessing I might have messed up in the
[MATH] x - 2y = 1 [/MATH] somewhere

 

[Deleted member 4993]

Hey !
So I have the following system :

[MATH] x - 2y = 1 [/MATH][MATH] ln(x) + ln(y) = 3ln(5) - ln(10) + ln(4) [/MATH]
So I'll explain what I tried first :
First attempt :
Figured out that x and y can't be negatives if I remember the lessons correctly.
Then, I got rid of the ln function and got this :
[MATH] x + y = 5^3 - 10 + 4 [/MATH]
With the new system, I got the following answer :
[MATH] y = 118/3 [/MATH][MATH] x = 239/3 [/MATH]But that's wrong.

On my second attempt I tried this :
[MATH] ln(xy) = ln(5^3/10)(4) [/MATH]followed by
[MATH] xy = 50 [/MATH]
With the new system, I got the following answer :
[MATH] y = 49/3 [/MATH]But I didn't even bother to continue this as those are the approximate forms I should be getting (They're on the screenshot I sent ) :

Can someone tell me what I'm doing wrong ? I wish I could focus a bit more on this exercice and figure it out myself but I got a lot to study for tomorrow and I can't stay stuck on this too long.

x * y = 50 → y = 50/x
and
x - 2y = 1

x - 2*(50/x) = 1.......... continue

 

[Steven G]

How did you conclude that x + y = -1/2? Maybe by dividing by -2? You can NOT divide the whole right side by -2 and only part of the left side by -2. What happened to dividing the x by -2 as well??

 

[Sap67]

I think this is good ! So thanks a lot for your help, all of you.

Jomo I think I was fixated on making the Sum and Product quadratic equation work that I tried my best to make it work, without thinking it through and realizing that it doesn't work here. At least not in an easy way.

 

[JeffM]

[MATH]ln(x) + ln(y) = 3ln(5) - ln(10) + ln(4) \implies ln(xy) = 3ln(5) - ln(10) + ln(4) =[/MATH]
[MATH]ln(5^3) - ln(5 * 2) + ln(2^2) = ln \left (\dfrac{5^3 * 2^2}{5 * 2} \right ) = ln (25 * 2) = ln(50).[/MATH]
Now you can use this law of logs

[MATH]\text {Given } a > 0 < b \ log_c(a) = log_c(b) \iff a = b.[/MATH]
One log on each side of the equality. You cannot remove logs willy-nilly before then.

[MATH]\therefore ln(xy) = ln(50) \implies xy = 50.[/MATH]
Straight forward if you remember to simplify your logarithmic expressions before trying to remove the log functions.

[MATH]\text {And } x - 2y = 1 \implies x = 2y + 1 \implies 50 = (2y + 1)y = 2y^2 + y \implies[/MATH]
[MATH]2y^2 + y - 50 = 0.[/MATH]
Now you were ABSOLUTELY headed in the right direction. If you will excuse my being fussy, however, I point out that it is inadvisable to approximate before the very end, and it is wrong to use equal signs when you mean approximation signs.

[MATH]1^2 - 4(2)(-50) = 1 + 400 = 401 > 400 \implies \sqrt{401} > \sqrt{400} = 20.[/MATH]
And you correctly saw that

[MATH]\dfrac{-1 - \sqrt{401}}{2 * 2} < 0 \implies y = \dfrac{-1 + \sqrt{401}}{4} \implies[/MATH]
[MATH]x = 2y + 1 = \dfrac{1 + \sqrt{401}}{2}.[/MATH]
That is the exact answer. Now, if you have been told to approximate to two decimals, you write

[MATH]y \approx 4.76 \text { and } x \approx 10.51.[/MATH]
Notice that you may not be able to validate your answer with approximations, but you can with exact answers.

[MATH]\dfrac{1 + \sqrt{401}}{2} - \dfrac{2(-1 + \sqrt{401})}{4} = \dfrac{1 + \sqrt{401}}{2} -\dfrac{-1 + \sqrt{401}}{2} = \dfrac{2}{2} = 1. \checkmark[/MATH]
[MATH]ln(x) + ln(y) = ln(xy) = ln \left ( \dfrac{1 + \sqrt{401}}{2} * \dfrac{-1 + \sqrt{401}}{4} \right ) =\\ ln \left( \dfrac{-1 + \sqrt{401} - \sqrt{401} + 401}{8} \right )= \ln \left ( \dfrac{400}{8} \right ) =\\ \ln \left (4 * \dfrac{100}{8} \right ) = \ln \left ( 4 * \dfrac{1}{10} + \dfrac{100}{0.8} \right ) = \ln \left ( 4 * \dfrac{1}{10} + 125 \right ) =\\ ln(125) - ln(10) + ln(4) = ln(5^3) - ln(10) + ln(4) = 3 ln(5) - ln(10) + ln(4). \checkmark[/MATH]