How do you solve #7^x=49#?

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Answer 1 · 2016-09-30T01:13:43

#x=2#

#7^x=49#
#=> 7^x=7^2#
#=> x=2#

Answer 2 · 2016-09-30T01:20:56

#x=2#

By observation we can see that #x=2# because #7^2=49#. But let's solve this another way. We can do it using logs.

#7^x=49#

#log7^x=log49#

#xlog7=log(7xx7)#

#x=log(7xx7)/log7#

#x=(log7+log7)/log7#

#x=log7/log7+log7/log7=1+1=2#