Question

$$( {\bar{x}} + \mu_{\bar{x}} )+( \alpha + \mu ) \times \log ( 100 ) =$$

Answer

$$bar(x)+m*u_bar(x)+2*(a^2*l*p*h+m*u)$$

Solution


Remove parentheses.
\[bar(x)+mu_bar(x)+(alpha+mu)\log{100}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[bar(x)+mu_bar(x)+({a}^{2}lph+mu)\log{100}\]
The base of 10 is implied.
\[bar(x)+mu_bar(x)+({a}^{2}lph+mu)\log_{10}{100}\]
Use Definition of Common Logarithm: \({b}^{a}=x\) if and only if \(log_b(x)=a\)1. Ask: If \({10}^{x}=100\), what is x?2. Answer: \(2\).
\[bar(x)+mu_bar(x)+({a}^{2}lph+mu)\times 2\]
Regroup terms.
\[bar(x)+mu_bar(x)+2({a}^{2}lph+mu)\]