How do you evaluate #log_343 (7)#?

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Answer 1 · 2016-09-23T02:54:21

#log_343(7)=1/3#

#log_343(7)=x#

Step 1: Rewrite as an exponential.
#343^x=7#

Step 2:Take the log of both sides.
#log343^x=log7#

Step 3: Use the log rule #loga^x=xloga#.
#xlog343=log7#

#x=log7/log343color(white)(aaaa)#Use a calculator
#x=1/3#

OR

Use the change of base formula #log_ba=loga/logb#
#log_343(7)=log7/log343=1/3color(white)(aaa)# Use a calculator

OR

Rewrite as an exponential and consider the powers of 7.
#343^x=7#

#343=7^3# or #root(3)343=7# or #343^(1/3)=7#
#x=1/3#