Answer by Theo(13342) (Show Source): You can put this solution on YOUR website! your problem can be written as:
log(2,(m/n)^7)
that means log of (m/n)^7 taken to the base of 2.
this is equivalent to:
7 * log(2,(m/n))
this is equivalent to:
7 * (log(2,m) - log(2,n))
properties of logs that are used are:
log(x^a) = a * log(x)
log(x/y) = log(x) - log(y)
the order of operations is important.
first you had to resolve log(x^a).
then you had to resolve log(x/y).
your original equation is:
log(2,(m/n)^7)
your final equation is:
7 * (log(2,m) - log(2,n))
CONFIRMATION OF RESULTS FOLLOWS:
by the basic definition of logarithms:
log(2,(m/n)^7) = y if and only if:
2^y = (m/n)^7
let's assume that m is equal to 10 and n is equal to 2.
the equation then becomes:
2^y = (10/2)^7 = 5^7 = 78125
we can use logs to solve this problem.
take the log of both sides of the equation to get:
log(2^y) = log(78125)
this simplifies to:
y*log(2) = log(78125) which further simplifies to:
y = log(78125)/log(2) which then becomes:
y = 16.25349666
when y = 16.25349666 and m = 10 and n = 2, we get:
2^y = (m/n)^7 which becomes:
2^16.25349666 = 5^7 which becomes:
78125 = 78125 which confirms the value of y = 16.25349666 when the value of m is equal to 10 and the value of n is equal to 2.
now we can confirm that our original equation and our final equation are the same.
our original equation is:
log(2,(m/n)^7)
replacing m with 10 and n with 2, we get:
log(2,((10/2)^7)) which becomes:
log(2,(5^7)) which becomes:
log(2,78125) which becomes:
16.25349666
our final equation is:
7 * (log(2,m) - log(2,n))
replacing m with 10 and n with 2, we get:
7 * (log(2,10) - log(2,2)) which becomes:
7 * (3.321928095 - 1) which becomes:
7 * (2.321928095) which becomes:
16.25349666
The answers are the same so the conversion process was done correctly.
To use your calculator to get the log of a number to a base other than 10, then do the following:
log(2,x) = LOG(x) / LOG(2)
LOG(x) means log(10,x) which means the log of x to the base of 10 which the calculator does automatically when you use the LOG function of the calculator.
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