The second equation can be rewritten as log(x-2y) = 2^5
The first equation is more complicated because you have different bases. You need to chane base 4 to base 2 by using the rule that allows for base changes.
Once you do this, you can use the Law of Logs to combine the left side which will equal 2^4 or 16.
Can you finish it now? Do you need more help?
solve the simultaneous equations;
log x base 2-log y base 4=4,
log (x-2y)base 2=5
2 answers
I have a strong feeling that Jessica has a typo , and all the bases are base 2
(assume when I write log (..) , I mean log2 (..) )
from the 1st:
log x - log y = 4 ----> x/y = 16 or x = 16y
from the 2nd:
log(x-2y) = 5 ---> x-2y =32
use substitution:
16 - 2y = 32
14y = 32
y = 16/7 , then x = 16(16/7) = 256/7
(256/7 , 16/7) satisfies both "assumed" equations)
(assume when I write log (..) , I mean log2 (..) )
from the 1st:
log x - log y = 4 ----> x/y = 16 or x = 16y
from the 2nd:
log(x-2y) = 5 ---> x-2y =32
use substitution:
16 - 2y = 32
14y = 32
y = 16/7 , then x = 16(16/7) = 256/7
(256/7 , 16/7) satisfies both "assumed" equations)
6 answers