Homework Soursesuppose that log7 x3 and log7 y4 use properties of logarithm
suppose that log^7 (x)=3 and log^7 (y)=4. use properties of logarithms to evaluate the following expression.
log^7(xy)
log^7(x/y^2)
log^7(x^2 and squar root of y/7)
log^7(49x^3y)
log^7(squar root of x/7y^3)
Help need to know how to solve
Solution
log^7 (x)=3 and log^7 (y)=4
loga^m = m loga
1). log^7(xy) = log^7(x) + log^7(y) [ by the formule logab = loga+logb]
=3 +4
=7
2). log^7(x/y^2) = log^7 (x) - log^7 (y)^2 [ by using fromule log a/b = log a - log b]
= 3 - 2 log^7(y) [ by log a^m = m loga ]
= 3 - 2.4
= -5
3) log^7 (x^2 . sqrt(y/7)
log^7 (x^2) + log^7 (y/7)
2log^7 x + log^7 (y) - log^7 7
2(3) +4 - 5.88
10 -5.88
4.12
4). log^7(49x^3 y)
log^7 (49) + 3 log^7(x) + log ^7 y [ by log abc = loga + logb +logc]
14 log7 + 3*3 + 4
11.83 +13 =24.83
5).
log^7(squar root of x/7y^3)
log^7 sqrt(x) - log^7 sqrt(7y^3)
1/2 (3) - 1/2 [ log ^7 7 + 3 log^7 y]
3/2 -1/2 [5.88 + 3*4]
3/2 - 1/2[17.88]
1.5 -8.94
-7.44
