How do you solve #2\log _ { 7} x = - \log _ { 7} 49#?

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Answer 1 · 2016-11-25T22:58:21

#x=1/7#

Use the rule #c*log_a(b)=log_a(b^c)# to rewrite both of these logarithms.

#2*log_7(x)=-1*log_7(49)#

#log_7(x^2)=log_7(49^-1)#

Since #49^-1=1/49#:

#log_7(x^2)=log_7(1/49)#

Since the logarithms are equal, so will the things inside of them:

#x^2=1/49#

Taking the square root of both sides:

#x=sqrt(1/49)=1/sqrt49=1/7#

Note that #x=-1/7# is not an answer because the contents of a logarithm must be positive.