Question

use properties of logarithms to expand the logarithmic expression below as much as possible.\\(\log_d \frac{\sqrt{x} y^6}{z^8}\\)\\(\log_d \frac{\sqrt{x} y^6}{z^8} = \square\\) (simplify your answer.)

Explanation:

Step1: Apply Quotient Rule

The quotient rule of logarithms states that $\log_b \frac{M}{N} = \log_b M - \log_b N$. So, for $\log_d \frac{\sqrt{x}y^6}{z^8}$, we can split it as $\log_d (\sqrt{x}y^6) - \log_d (z^8)$.

Step2: Apply Product Rule to $\log_d (\sqrt{x}y^6)$

The product rule of logarithms is $\log_b (MN) = \log_b M + \log_b N$. Applying this to $\log_d (\sqrt{x}y^6)$, we get $\log_d \sqrt{x} + \log_d y^6$.

Step3: Simplify $\log_d \sqrt{x}$

Since $\sqrt{x} = x^{\frac{1}{2}}$, using the power rule $\log_b M^n = n\log_b M$, we have $\log_d x^{\frac{1}{2}} = \frac{1}{2}\log_d x$.

Step4: Simplify $\log_d y^6$

Using the power rule, $\log_d y^6 = 6\log_d y$.

Step5: Simplify $\log_d z^8$

Using the power rule, $\log_d z^8 = 8\log_d z$.

Step6: Combine All Terms

Putting it all together: $\frac{1}{2}\log_d x + 6\log_d y - 8\log_d z$.

Answer:

$\frac{1}{2}\log_{d} x + 6\log_{d} y - 8\log_{d} z$