Logs simultaneous equation

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

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?

Consider what would happen if b were negative, can you take the logarithm of a negative number?

Ah okay I guess not then

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

It's quite possible. $\log(-x)$ exists for all real $x < 0$.