How do you solve #7^x=1/49#?

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Answer 1 · 2017-11-20T10:06:16

By arranging the equation

#7^x = 1/49 = 1/(7^2)#

#(7^x)times(7^2)=1#

zeroth power of any number is 1.

#x+2 = 0#

#x=-2#

This is your answer #x=-2#