Given that log x= 5 and log y= -7. Evaluate the following :
a) log xy.
b) log 1/x.
c) log y/x.
d) log xy^2.
a
log x y = log x + log y = -2
b
10^log x = x
10^0 = 1
so
log 0 = 1
so now the problem
log 1/x = log 1 - log x = 0 - log x = -log x = -5
c.
log y - log x obviously
-7 - 5 = - 12
d.
log x + 2 log y
5 + 2(-7) = 5 - 14 = -9
LOL
I mean log 1 = 0 !!!!!!
Thanks
You are welcome .
To evaluate the expressions, we can make use of the following logarithmic properties:
1. Product Rule: log(ab) = log(a) + log(b)
2. Quotient Rule: log(a/b) = log(a) - log(b)
3. Power Rule: log(a^b) = b * log(a)
Let's apply these rules to evaluate each expression:
a) log(xy):
Using the product rule, we have: log(xy) = log(x) + log(y)
Substituting the given values: log(xy) = log(x) + log(y) = 5 + (-7) = -2
b) log(1/x):
We can rewrite 1/x as x^(-1).
Using the power rule, we have: log(1/x) = log(x^(-1)) = -1 * log(x)
Substituting the given value: log(1/x) = -1 * log(x) = -1 * 5 = -5
c) log(y/x):
Using the quotient rule, we have: log(y/x) = log(y) - log(x)
Substituting the given values: log(y/x) = log(y) - log(x) = -7 - 5 = -12
d) log(xy^2):
Using the product and power rules, we have: log(xy^2) = log(x) + log(y^2) = log(x) + 2 * log(y)
Substituting the given values: log(xy^2) = log(x) + 2 * log(y) = 5 + 2 * (-7) = 5 - 14 = -9
Therefore,
a) log(xy) = -2
b) log(1/x) = -5
c) log(y/x) = -12
d) log(xy^2) = -9