How do you evaluate #log_49 343#?

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Answer 1 · 2018-05-25T11:52:59

#log_49 343 = 3/2#

#log_49 343#
#=log_7(343)/log_7(49)#
#=log_7(7^3)/log_7(7^2)#
#=(3log_7(7))/(2log_7(7))#
#=3/2#

Alternatively, you could say #343 = 49 ^ (3/2)#

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Answer 2 · 2018-05-26T08:36:45

#log_(49)343=3/2#

#x=log_(49)343#

#=>49^x=343#

#=>(7^2)^x=343#

#7^(2x)=343#

but #7^3=343#

#:.2x=3#

#=>x=3/2#

#log_(49)343=3/2#