How do you solve #log_10 7=log_10 x-log_10 2#?

1 Answer
Tony B · zoe
Oct 10, 2015

#x=14#

Explanation:

For two logarithms of the same base, #log_a M=log_a N rArr M=N#

Using the Quotient Law, #log_a x-log_a y=log_a (x/y)#

So,

#log_10 7= log_10 (x/2)#
#7=x/2#
#x=7*2#
#x=14#

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ABR May I suggest another approach:

Logs are indices in another form as such adding indices is another representation of multiplication. Like wise, subtraction is another representation of division. Consequently:

It is given that #log7 = log x - log 2#

( it does not matter a bout what the base is in this context)

#Antilog( log 7) = 7, Antilog( log x) =x, Antilog( log 2) =2#

Thus #log 7 = log x - log 2 " is equivalent to " 7 = x/2#

From this it is a simple matter of manipulation to find #x#

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