Question

log_f \frac{\sqrt{s} t^6}{r^2} = \square (simplify your answer.)

Explanation:

Step1: Apply Logarithm Quotient Rule

The logarithm quotient rule states that \(\log_b\frac{M}{N}=\log_b M - \log_b N\). So, for \(\log_f\frac{\sqrt{s}t^6}{r^2}\), we can split it as \(\log_f(\sqrt{s}t^6)-\log_f(r^2)\).

Step2: Apply Logarithm Product Rule

The logarithm product rule is \(\log_b(MN)=\log_b M+\log_b N\). Applying this to \(\log_f(\sqrt{s}t^6)\), we get \(\log_f\sqrt{s}+\log_f t^6\). Now our expression is \(\log_f\sqrt{s}+\log_f t^6-\log_f r^2\).

Step3: Rewrite Radicals as Exponents

Recall that \(\sqrt{s}=s^{\frac{1}{2}}\). So, \(\log_f\sqrt{s}=\log_f s^{\frac{1}{2}}\).

Step4: Apply Logarithm Power Rule

The logarithm power rule is \(\log_b M^n = n\log_b M\). Applying this to each term:

• For \(\log_f s^{\frac{1}{2}}\), we get \(\frac{1}{2}\log_f s\).
• For \(\log_f t^6\), we get \(6\log_f t\).
• For \(\log_f r^2\), we get \(2\log_f r\).

Putting it all together, the simplified expression is \(\frac{1}{2}\log_f s + 6\log_f t- 2\log_f r\).

Answer:

\(\frac{1}{2}\log_{f} s + 6\log_{f} t - 2\log_{f} r\)