Calculate the value of #p# in #log49^2=plog7# ?

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Answer 1 · 2017-06-23T08:18:46

#p = 4#

#log 49^2=p log 7 rArr log(7^2)^2=log 7^p# or

#log 7^4= log 7^p rArr p = 4#

Answer 2 · 2017-06-23T08:20:59

#p=4#

#Log49^2 = pLog7#

Solution

Recall #-># #aLogb = Logb^a#

#:.# #Log49^2 = Log7^p#

#Log7^(2(2)) = Log7^p#

#Log7^4 = Log7^p#

#cancel(Log7)^4 = cancel(Log7)^p#

Hence #4=p#

#p=4# #-># #Answer#

Answer 3 · 2017-06-23T08:27:08

Use the following log law:

#logx^y=ylogx#

and the following index law:

#(x^a)^b=x^(ab)#

Simplify as follows:

#log49^2=plog7#

#rArrlog(7^2)^2=plog7#

#rArrlog7^4=plog7#

#rArr4log7=plog7#

#rArrp=4cancel(log7)/cancel(log7)#

#rArrp=4#