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How do you write #log_7 (1/49)=-2# in exponential form?

1 Answer

Jan 27, 2017

see explaination

Explanation:

Let say #log_7(1/49)=y#

#(1/49) = 7^y# use concept if #log_ab=c#, then #b=a^c#

#1/7^2=7^y#

#7^-2=7^y#

#y=-2#

therefore, #log_7(1/49)=-2#

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