Solve an equation containing a log?
How do you solve this equation?
log(3x+1)=2
That is how the equation is listed, so I would say its safe to assume its log base 10.
How do you solve this equation?
log(3x+1)=2
That is how the equation is listed, so I would say its safe to assume its log base 10.
Raymond
Favorite Answer
Find the "antilog"
If the log is in base 10, then the "antilog" of 2 is 10^2 (equal to 100)
the antilog of a log is simply the "argument" of the log.
(if the base is 10)
10^(log(3x+1)) = 3x+1
Therefore, by taking the antilog on both sides, you would get
3x + 1 = 10^2
(if the base is not 10, then simply use the correct base):
3x + 1 = (base)^2
Ed
Assuming you mean log to the base of 10,
3x + 1 = 10^2 = 100 so 3x = 99 so x = 33
lenpol7
log(3x+1)=2
Antilog
3x + 1 = 10^2
3x + 1 = 100
3x = 99
x = 33
Professor Of Wumbology
log(b) = a just means 10^a = b.
"log(3x+1) = 2" turns into "10^2 = 3x+1"
10^2 = 3x + 1
100 = 3x + 1
99 = 3x
x = 33
Chris
antilog 2 = 100
Hence log 100 = 2
Hence 3x + 1 = 100
Hence 100 - 1 = 99 = 3x
Therefore x = 99/3 = 33