Question

evaluate the expression.
\log_{0.2} 11.3 \approx \square

Explanation:

Step1: Recall the change of base formula

The change of base formula for logarithms is \(\log_{a}b=\frac{\log_{c}b}{\log_{c}a}\), where \(c>0,c
eq1\). We can use the common logarithm (base 10) or the natural logarithm (base \(e\)). Let's use the natural logarithm here. So \(\log_{0.2}11.3 = \frac{\ln(11.3)}{\ln(0.2)}\).

Step2: Calculate the natural logarithms

First, calculate \(\ln(11.3)\approx2.4239\) (using a calculator to find the natural logarithm of 11.3). Then, calculate \(\ln(0.2)\approx - 1.6094\) (using a calculator to find the natural logarithm of 0.2).

Step3: Divide the two results

Now, divide \(\ln(11.3)\) by \(\ln(0.2)\): \(\frac{2.4239}{-1.6094}\approx - 1.506\).

Answer:

\(\approx - 1.51\) (rounded to two decimal places, or more precise value as per calculation)