Question 1185384: If logb(2)=1.2 and logb(3)=1.5, evaluate logb(6)=?
Found 3 solutions by ikleyn, Theo, MathTherapy:
Answer by ikleyn(52794)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! one of the log rules says that log(a * b) = log(a) + log(b).
this rule applies regardless of the base, as long as the base is the same for all operands.
you can use your calculator to confirm this is true.
the log function of your calculator works with the base of 10.
the ln function of your calculator works with the base of e.
assume your function is log5(25).
since log5(25) = y if and only if 5^y = 25, then you know that y must be equal to 2 because 5^2 = 25.
using the base conversion formula to convert to the log function of the calculator and the ln function of the calculator, you get:
log5(25) = log(25)/log(5) = 2 and log5(25) = ln(25)/ln(2) = 2.
the log conversion formula works for any log base.
you can use this log base conversion formula to confirm that log(a * b) = log(a) + log(b).
assume a = 15 and b = 20.
a * b = 15 * 20 = 300
log(15 * 20) = log(300) = 2.477121255 if you're working in the log base of 10.
log(15) + log(20) = the same.
ln(15 * 20) = ln(300) = 5.703782475 if you're working in the log base of e.
ln(15) + ln(20) = the same.
if you're working in the base of b, you can't use you're calculator to confirm it's true, but you can be assured that the property of logs applies with the base b as well.
logb(2) = 1.2
logb(3) = 1.5
logb(2 * 3) = logb(2) + logb(3) = 1.2 + 1.5 = 2.7
since 2 * 3 = 6, then logb(6) = 2.7
since the property works with all bases, than assume b = 10 and you get:
log(2) = .3010299957
log(3) = .4771212547
log(2 * 3) = log(6) = .7781512504
log(2) + log(3) = the same.
your solution is that logb(6) = 1.2 and 1.5 = 2.7.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website! If logb(2)=1.2 and logb(3)=1.5, evaluate logb(6)=?
logb (2) = 1.2 logb (3) = 1.5
logb (6) = logb (3 * 2) = logb (3) + logb (2) = 1.5 + 1.2 = 2.7
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