How do you evaluate #log 0.01 #?

3 Answers
Gió
Mar 22, 2016

I found #-2# if the log is in base #10#.

Explanation:

I would imagine the log base being #10#
so we write:
#log_(10)(0.01)=x#
we use the definition of log to write:
#10^x=0.01#
but #0.01# can be written as: #10^-2# (corresponding to #1/100#).
so we get:
#10^x=10^-2#
to be equal we need that:
#x=-2#
so:
#log_(10)(0.01)=-2#

Bdub
Mar 22, 2016

#log 0.01=-2#

Explanation:

#log 0.01#
#=log (1/100)#
#=log(1/10^2)#
#=log10^-2#-> use property #1/x^n = x^-n#
#-2log10#->use property #log_b x^n=n*log_bx#
# = -2(1)#->log 10 is 1
#=-2#

Harish Chandra Rajpoot
Jul 21, 2018

#-2#

Explanation:

#\log0.01#

#=\log(1/100)#

#=\log(10^{-2})#

#=-2\log10#

#=-2\cdot 1#

#=-2#

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