If 5^x=3 and 9^y=125, find the value of xy.
For the first equation
5x = 3
Take the log of both sides
log(5x) = log(3)
Use the rule of logs on LHS: logAB = B·logA
x·log5 = log3
Divide both sides by log5
log3
x = ————
log5
====
For the second equation
9y = 125
Take the log of both sides
log(9y) = log125
As before, use the rule of logs on LHS: logAB = B·logA
y·log9 = log125
Divide both sides by log9
log125
y = ——————
log9
log3 log125
So xy = ———— · ——————
log5 log9
But we can simplify that
log3 log125
———— · ——————
log5 log9
write 125 as 53 and 9 as 32
log3 log53
———— · ——————
log5 log32
Again we use the rule logAB = B·logA
log3 3log5
———— · ——————
log5 2log3
Cancel the log3's
1
log3 3log5
———— · ——————
log5 2log3
1
Cancel the log5's
1 1
log3 3log5
———— · ——————
log5 2log3
1 1
All that's left is
3
———
2
Edwin
AnlytcPhil@aol.com