Log5 (log3x) =0?

Question

Log5 (log3x) =0?

Answer 1

(log 5)(log 3x)=0

Divide out log 5

log 3x=0/log 5

log 3x=0

Change out of log form

10^0=3x

(10^0 = 1)

Then

1=3x

Final Answer

x=1/3

Or you can do it this way

(log 5)(log 3x)=0

Convert From logs

(5^0)(10^0)=3x

1=3x

x= 1/3

Answer 2

Lets assume that the 5 in your sum indicates base of the log.

If so then we know that Log of any quantity [ in this case it is (log3x)] to any base is zero only when the quantity of which the log is being taken = 1. Here in our case - Log of (log3x) to the base 5 = 0

ie log 3x must be equal to 1.

=>  log 3x  to any base = 1 ----------

Now two cases arise -

Case 1 --- Log base is 10. It means that  log of 3x to the base 10 = 1

=> log 3x  = 1 = log 10

=> 3x = 10

=> x = 10/3 .................... Answer

Case 2 .....  If the base is e then

log 3x = 1 = log e

=> 3x = e

=>   x = e/3 ..................  Answer

Still if you choose 5 as the base then -

log 3x = 1 = log 5

=> 3x = 5

=>   x = 5/3 ..................... Answer

--------------------------------------...

Note Answer will vary with the change in log base.

Answer 3

andrej m...

i just try it ok?

log5(log3x)=0

log5+log3x=0

log(5+3x)=0

5+3x=10^0

5+3x=1

     3x=1-5

     3x=-4

       x=-4/3

Answer 4

log 5(log3x)=0

log3x= 0/log5

log3x=0

log(base 10)(3x)= 0

10^0 = 3x

x=1/3

Answer 5

Is this log to the base 5 of the log of 3x = 0?

If so, convert it to an exponential equation: 5 ^ 0 = log (3x)

5^0 = 1, and the log of 10 is 1, so 3x = 10

and x = 10/3