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# Logs simultaneous equation

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1. Solve the simultaneous equations:

ab= 25

log4a - log4b = 3

How do you do this?
2. Write the second equation as log a - log b = log 64, then write the LHS as log (a/b) and remove the logs. You then have SEs without logs
3. Simultaneously solving Log4a-log4b=3 and ab=25
Can be written as: log4(a/b)=3
Removing the log:a/b=43 which is: a/b=64
Rearranging: a=64b
Substitute this into the second equation and solve for rest
4. (Original post by teamjudo)
Simultaneouslysolving Log4a-log4b=3andab=25Can be written as: log4(a/b)=3Removing the log:a/b=43 which is: a/b=64Rearranging: a=64bSubstitute this into the second equation: (64b)b=25Therefore: 64b2=25 So: b2=25/64And: b=sqrt(25/64) which is: b=5/8Again substitute this into the second equationfor a: a(5/8)=25Solve for a: a=25(8/5)Therefore: a=40
Full solutions (especially when they're horribly formatted) are against forum guidelines. Thanks.
5. (Original post by nerak99)
Write the second equation as log a - log b = log 64, then write the LHS as log (a/b) and remove the logs. You then have SEs without logs
Oooh thank you!!

I've gotten to b2 = 25/64 so would this give two solutions for b ( 5/8 and -5/8) or just one?
6. (Original post by jessyjellytot14)
Oooh thank you!!

I've gotten to b2 = 25/64 so would this give two solutions for b ( 5/8 and -5/8) or just one?
Consider what would happen if b were negative, can you take the logarithm of a negative number?
7. (Original post by Zacken)
Consider what would happen if b were negative, can you take the logarithm of a negative number?
Ah okay I guess not then
8. (Original post by jessyjellytot14)
Ah okay I guess not then
It's still good practice to write down both answers and then 'reject' one by giving a reason in an A-Level exam, sometimes marks are awarded for this.
9. (Original post by jessyjellytot14)
Oooh thank you!!

I've gotten to b2 = 25/64 so would this give two solutions for b ( 5/8 and -5/8) or just one?

It is not possible to have a log(-x) for example
10. (Original post by teamjudo)

It is not possible to have a log(-x) for example
It's quite possible. exists for all real .
11. (Original post by Zacken)
It's quite possible. exists for all real .
Well if we are noting that --x x>0 is positive then that is a waste of everyones time here.

But in Core maths, no log of a negative number and so that is a bit of a distraction.

In Core maths a log can be negative but you cant log a negative

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