Question

find the numerical value of the log expression.\
\\(\log a = -5\\) \\(\log b = 12\\) \\(\log c = -9\\)\
\\(\log \frac{c^7}{a^6b^2}\\)

Explanation:

Step1: Apply log quotient and product rules

Using \(\log\frac{M}{N}=\log M - \log N\) and \(\log(MN)=\log M+\log N\), we get \(\log\frac{c^{7}}{a^{6}b^{2}}=\log c^{7}-\log(a^{6}b^{2})=\log c^{7}-(\log a^{6}+\log b^{2})\).

Step2: Apply log power rule

Using \(\log M^{n}=n\log M\), we have \(\log c^{7}=7\log c\), \(\log a^{6}=6\log a\), \(\log b^{2}=2\log b\). So the expression becomes \(7\log c - 6\log a - 2\log b\).

Step3: Substitute given values

Substitute \(\log a = - 5\), \(\log b = 12\), \(\log c=-9\) into the expression: \(7\times(-9)-6\times(-5)-2\times12\).

Step4: Calculate each term

Calculate \(7\times(-9)=-63\), \(6\times(-5)= - 30\) (so \(-6\times(-5)=30\)), \(2\times12 = 24\) (so \(-2\times12=-24\)).

Step5: Sum the terms

Sum: \(-63 + 30-24=-57\).

Answer:

\(-57\)