How do you calculate #log_(343) 49#?

1 Answer
Alan P.
May 4, 2016

#log_343 49 = 2/3# (method below)

Explanation:

Notice that
#color(white)("XXX")49=7^2#
and
#color(white)("XXX")343=7^3#

and remember that #log_b a = k# is equivalent to #b^k=a#

Let #color(blue)(log_343 49=k)#

#color(white)("XXX")rArr 343^k=49#

#color(white)("XXX")rArr (7^3)^k=7^2#

#color(white)("XXX")rArr 7^(3k)=7^2#

#color(white)("XXX")rArr 3k=2#

#color(white)("XXX")rArr color(blue)(k=2/3)#

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