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

1 Answer
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

=================================
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