How do you write #log0.001=x# in exponential form?

2 Answers
anor277
Oct 26, 2015

If we're using base #10#, and I assume we are, #log# #(0.001)# #=# #-3#. Why?

Explanation:

When we write #log_ab = c#, we are asking to what power we raise the base #a# to get #c#. Here, #a^c = b#.

Here the base is #10#. So in fact we are asking to what power we raise #10# to get #0.001#. This is clearly #-3#, because #10^-3# #=# #1/10^3# #=# #1/1000# #=# #0.001#. Apologies if have not grasped your question.

EZ as pi
Sep 24, 2016

#log_10 0.001 = x " "hArr " "10^x= 0.001#

Explanation:

Log form and exponential form are two ways to say the same thing.
They are interchangeable..

#log_a b = c " "hArr" "a^c = b#

In this case we have

#log_10 0.001 = x " "hArr " "10^x= 0.001#

The question being asked in each form, is ....

"To what power must 10 be raised to give 0.001?"

To change from one to the other, remember..
"The base stays the base and the other two change around"

The answer for x can be found from:

#10^x = 0.001 = 1 xx 10^-3#

#x = -3#

#log_10 0.001 = -3 " " or " "10^-3 = 0.001#

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