We are given the following expression:
[tex]\log \frac{\sqrt[3]{b^2c^4}}{a^8}[/tex]To determine the numerical value of the expression we will use different "log" properties to expand it.
First, we will use the following property:
[tex]\log \frac{x}{y}=\log x-\log y[/tex]Applying the property we get:
[tex]\log \frac{\sqrt[3]{b^2c^4}}{a^8}=\log \sqrt[3]{b^2c^4}-\log a^8[/tex]Now, we will use the following property of roots:
[tex]\sqrt[n]{x}=x^{\frac{1}{n}}[/tex]Applying the property we get:
[tex]\log \sqrt[3]{b^2c^4}-\log a^8=\log (b^2c^4)^{\frac{1}{3}}-\log a^8[/tex]Now, we use the following property:
[tex]\log x^n=n\log x[/tex]Applying the property:
[tex]\log (b^2c^4)^{\frac{1}{3}}-\log a^8=\frac{1}{3}\log b^2c^4-8\log a[/tex]Now, we apply the following property on the "log" on the left side:
[tex]\frac{1}{3}\log b^2c^4-8\log a=\frac{1}{3}\log (b^2)+\frac{1}{3}\log c^4-8\log a[/tex]Now, we apply the property of exponents:
[tex]\frac{1}{3}\log (b^2)+\frac{1}{3}\log c^4-8\log a=\frac{2}{3}\log (b^{})+\frac{4}{3}\log c^{}-8\log a[/tex]Now, we substitute the given values for each of the "log":
[tex]\frac{2}{3}\log (b^{})+\frac{4}{3}\log c^{}-8\log a=\frac{2}{3}(-9)+\frac{4}{3}(-9)-8(-10)[/tex]Solving the operations:
[tex]\frac{2}{3}(-9)+\frac{4}{3}(-9)-8(-10)=62[/tex]Therefore, the numerical value of the expression is 62.
