Verified answer

Yes, you are correct.

One way to do log problems is to remember the connection between the following two equations:

x=b^y and y=log_b (x)

They are equivalent.

So one way to solve this question is:

log x = 0.00001

log x = y form, with base 10

change it to x=b^y form

you have x=10^0.00001

which is very close to 1, since 0.00001 is very close to 0.

x = 10^0.00001

is correct

Correct.

whenever you spot 2 numbers that are being mulitplied you need to use log(ab) = loga+logb so log15 = log3+log5 = log1+log15 and so on.... in case you spot branch of two quantity then its SUBSTRACTION in case you spot skill then its the final one...... Multiply ===> Addition in log international Divison ==> Subtraction in Log international