Question

4logx-3log(X^2+1)+36log(x-1)
rewrite the expression as a single logarithm log A. then the function
A=

Answer 1

The coefficient of a logarithmic expression can be rewritten as the exponent of the logarithm's interior expression. For example:
\[4 \log\ x = \log\ x ^{4}\]
\[3 \log \left( x^2+1 \right) = \log \left( x^2+1 \right) ^{3}\]
\[36 \log \left( x-1 \right) = \log \left( x - 1 \right)^{36}\]
In addition, the operators between logarithmic expression can allow them to merge their interior expressions. Addition translates to multiplication interior expressions, and subtraction translates to division of interior expressions. For example:
\[ \log\ x ^{4} + \log \left( x^2+1 \right) ^{3} = \log\ x ^{4} \left( x^2+1 \right) ^{3}\]
\[\log\ x ^{4} \left( x^2+1 \right) ^{3} - \log \left( x - 1 \right)^{36} = \log\ \frac{ x ^{4} \left( x^2+1 \right) ^{3} }{ \left( x - 1 \right)^{36} }\]
The A value your prompt asked for is what's inside this logarithmic expression. (You may simplify it further if you wish.)