This is a logs question
If u=x/y^2, which expression is equivalent to log u?
1) log x + 2 log y
2) 2(log x- log y)
3) 2(log x + log y)
4) log x- 2 log y
log(x/y^2) = log(x) - log(y^2)
so, (4)
Well, well, well, it looks like we have a fun logarithmic question here!
To find the expression equivalent to log u, we can use the logarithmic property that says log A - log B = log(A/B).
In our case, u = x/y^2. We can rewrite it as u = x(y^-2).
Using the logarithmic property, we can hide our clown noses and conclude that log u = log(x(y^-2))
Which brings us to the stunning revelation that the correct expression equivalent to log u is option 4) log x - 2 log y!
Now, go forth in logarithmic merriment, my curious friend!
To find the expression equivalent to log u, we can use the logarithm properties. The basic logarithm property states that log a - log b = log (a/b).
Here we have u = x/y^2. Now, let's express u in terms of logs:
u = x/y^2
Taking the log base 10 of both sides:
log u = log (x/y^2)
Now, using the logarithm property, we can rewrite this expression:
log u = log x - log y^2
Next, we use another logarithm property, log a^b = b log a, to simplify further:
log u = log x - 2 log y
So, the expression equivalent to log u is:
log u = log x - 2 log y
Therefore, option 4) log x - 2 log y is the correct choice.
To determine which expression is equivalent to log u when u = x/y^2, we can use the properties of logarithms. One of the key properties states that log(a/b) is equal to log a - log b.
Starting with the given expression, u = x/y^2, we can apply this property to obtain:
log u = log(x/y^2) = log x - log y^2
Next, we apply another property of logarithms which states that log(a^b) is equal to b * log a. In this case, we have log y^2, which can be rewritten as 2 * log y:
log u = log x - 2 * log y
Comparing this expression to the options given:
1) log x + 2 log y
2) 2(log x - log y)
3) 2(log x + log y)
4) log x - 2 log y
We can see that option 4, log x - 2 log y, is equivalent to log u based on the properties of logarithms we applied.