Solve an equation containing a log?

Question

How do you solve this equation?

log(3x+1)=2

Raymond · Accepted 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 · Answer

Assuming you mean log to the base of 10, 

3x + 1 = 10^2 = 100 so 3x = 99 so x = 33

lenpol7 · Answer

log(3x+1)=2

Antilog 

3x + 1 = 10^2 
3x + 1 = 100 
3x = 99 
x = 33

Professor Of Wumbology · Answer

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 · Answer

antilog 2 = 100

Hence log 100 = 2
Hence 3x + 1 = 100
Hence 100 - 1 = 99 = 3x
Therefore x = 99/3 = 33

Rita the dog · Answer

3x+1 = 10^2, so 3x=99 and x=33.

Solve an equation containing a log?

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