how do you get the (x^7/2) from
1/2 loga X + 3 loga Y - 4 loga X
assuming all logs are base a, and fixing your typo,
1/2 logx + 3logx - 4logx
= log x^(1/2) + log x^3 - log x^4
= log (x^(1/2) * x^3 / x^4)
= log x^(1/2 + 3 - 4)
= log x^(-1/2)
Clearly that is not all that's wrong.
Go look at the problem again, and recall that
log x^n = n logx
log(x*y) = logx + logy
To solve this expression, you need to use logarithm properties. Let's break it down step by step:
1. First, recall the logarithm property: logₐ(b) + logₐ(c) = logₐ(b * c). This means that if we have two logarithms with the same base, we can combine them by multiplying their arguments.
2. In the given expression, we have 1/2 logₐ(X), 3 logₐ(Y), and -4 logₐ(X). We can rewrite these as:
1/2 logₐ(X) = logₐ(X^(1/2))
3 logₐ(Y) = logₐ(Y^3)
-4 logₐ(X) = logₐ(X^(-4))
3. Next, we apply the logarithm property from step 1 to combine all the logarithms with the same base, a:
logₐ(X^(1/2)) + logₐ(Y^3) + logₐ(X^(-4))
4. Another logarithm property states that logₐ(b) + logₐ(c) = logₐ(b * c), so we can multiply the arguments:
logₐ(X^(1/2) * Y^3 * X^(-4))
5. Now, let's simplify the expression inside the logarithm:
X^(1/2) * Y^3 * X^(-4)
To multiply exponential expressions, we can add their exponents:
X^(1/2 - 4) * Y^3
X^(-7/2) * Y^3
6. Finally, we substitute the simplified expression back into the logarithm:
logₐ(X^(-7/2) * Y^3)
Using the power rule of logarithms, we can bring down the exponent:
(-7/2) logₐ(X) + 3 logₐ(Y)
Thus, the expression 1/2 logₐ(X) + 3 logₐ(Y) - 4 logₐ(X) simplifies to (-7/2) logₐ(X) + 3 logₐ(Y).