Let log10 70=m and log10 20=p. Given that log10 14=Am+Bp+C, where A, B,and C are integers, compute the ordered triple (A, B, C).
log14 = log2 + log7 = log(20/10)+log(70/10) = log20-1 + log70-1 = m+p-2
To solve this problem, we need to use the properties of logarithms and apply them to the given equation.
First, let's use the property of logarithms that states log(a*b) = log(a) + log(b).
From the given information, we have:
log10 70 = m ..............(1)
log10 20 = p ..............(2)
Now, let's find a relationship between log10 70, log10 20, and log10 14.
Using the property log(a*b) = log(a) + log(b), we can say that:
log10 14 = log10 (2*7).
Now, applying the property to each factor, we get:
log10 (2*7) = log10 2 + log10 7.
Since we have log10 20 = p, we can write log10 2 as log10 (4*5), and apply the logarithmic property again:
log10 (4*5) = log10 4 + log10 5.
log10 (4*5) = 2*log10 2 + log10 5.
Now, substituting the values of log10 2 and log10 20 into the equation for log10 14:
log10 14 = 2*log10 2 + log10 5 + log10 7.
Finally, comparing this equation with the given equation log10 14 = Am + Bp + C, we can determine the values of A, B, and C:
A = 2
B = 1
C = log10 5 + log10 7.
Therefore, the ordered triple (A, B, C) is (2, 1, log10 5 + log10 7).
To compute the ordered triple (A, B, C), we need to relate log10 14 to the given values of m and p.
We can use logarithmic properties to rewrite log10 14 in terms of m and p. First, let's express 14 as a product of 70 and 20 raised to some powers.
14 = 70 * 20^x
Now, taking the logarithm of both sides:
log10 14 = log10 (70 * 20^x)
By the properties of logarithms, we can split this into two separate terms:
log10 14 = log10 70 + log10 (20^x)
Substituting the values of m and p, we have:
log10 14 = m + px
Now we need to find the values of A, B, and C. Comparing the terms, we can see that:
A = 1, B = x, and C = 0.
Hence, the ordered triple is (1, x, 0).