Log Q

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I'm having trouble with these simultaneous Log equations. Any help? I have got the first equation to look like xy=243. Not sure what to do with the second one.


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TheBody

The charter forbids us from doing it out for you but if you meet us half way, we can give LOTS of help.

I would probably advise leaving everything in its power form. It'll prob help.

For example, beginning with

[latex]\log_9xy=\frac{5}{2}[/latex]
[latex]\Rightarrow xy=(9)^{\frac{5}{2}}[/latex]
[latex]\Rightarrow xy=(3^2)^{\frac{5}{2}}[/latex]
[latex]\Rightarrow xy=3^5[/latex]
[latex]\Rightarrow y=\frac{3^5}{x}[/latex].

I'd then replace this into the second equation and solve for x.

Give it a go and see how you get on.

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Thanks. Here is what I have. The answers work for the first equation but I get errors when I try it on my calculator for the second one.

TheBody

Good effort!!

Your mistake is on the 4th line.

[latex](\log_3x)^2\not=\log_3x^2[/latex]

So on line 4 you really need:

[latex]5(\log_3x)-(\log_3x)^2=-6[/latex]

[latex]\Rightarrow(\log_3x)^2-5\log_3x-6=0[/latex]

[latex]\Rightarrow(\log_3x-6)(\log_3x+1)=0[/latex]

Can you finish it from there?

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Ah I see it now. Couldn't see that substitution before. I got x =729 and y=1/3.

TheBody

Yes, that is one solution. The other way around is also a solution. ie x=1/3 and y=729.



Also, 5,000 posts for me!! Do I get a medal or something???!!

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TheBody wrote: »

Yes, that is one solution. The other way around is also a solution. ie x=1/3 and y=729.



Also, 5,000 posts for me!! Do I get a medal or something???!!

Nice one. What a way to reach 5000 posts. Talking about logs! Thanks for the help.

TheBody

No problem. Glad I could help. Happy Christmas!!

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OK I have another one to start the new year off. I think I have (a) out I think. I got -0.005t and 497 days for those answers. Its (b) I am having trouble with. I have posted as far as I have got.

TheBody

Sorry for the delay in getting back to you. I'm up to my eyes in work!!

Recall the rules of logs:
[latex]\ln(xy)=\ln(x)+\ln(y)[/latex]
[latex]\ln(\frac{x}{y})=\ln(x)-\ln(y)[/latex]
[latex]\ln(x^y)=y\ln(x)[/latex]

Also, as the natural log and the expenontial function are inverses of each, other we also have:
[latex]\ln(e^x)=x[/latex]
[latex]e^{\ln(x)}=x[/latex]

Anyway, back to your problem. The next step is to talk the natural log of both sides. I'll begin with your last line.
[latex]e^{bt+bk}=\frac{1}{2}e^{bt}[/latex]
[latex]\Rightarrow \ln{(e^{bt+bk})}=\ln{(\frac{1}{2}e^{bt})}[/latex]
[latex]\Rightarrow bt+bk=\ln{(\frac{1}{2})+\ln(e^{bt})}[/latex]
[latex]\Rightarrow bt+bk=\ln(1)-\ln(2)+bt}[/latex]

Can you finish it from there?

Out of curiosity, what exam paper is this from?

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Thanks. I have ended up with k = (-1/b)ln2.

They are from Mock papers I am giving a relative a hand with. Logs were never my strong suit. :rolleyes:

TheBody

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Thanks. I have ended up with k = (-1/b)ln2.

They are from Mock papers I am giving a relative a hand with. Logs were never my strong suit. :rolleyes:

Well done. That's what I got too.