91Ó°ÊÓ

To rewrite the expression \(\log_4{6w}\) as a sum or difference of logarithms, we will need to use the logarithm properties, particularly the product rule and the change of base formula.

The product rule states that \(\log_b{MN} = \log_b{M} + \log_b{N}\), where b is the base, M and N are positive real numbers. In this case, we have a product inside the logarithm, 6w, so we can apply the product rule to the expression: \(\log_4{6w} = \log_4{6} + \log_4{w}\)

If you need to simplify the expression further by converting to the common logarithm (logarithm with base 10) or the natural logarithm (logarithm with base e), you can use the change of base formula. The change of base formula states that \(\log_b{M} = \frac{\log_c{M}}{\log_c{b}}\), where c can be any base other than the original one (most commonly 10 or e). Note that this step is optional and may not be required depending on your teacher's requirement or the context of the problem. However, we will demonstrate it for both terms in our expression using the natural logarithm. \(\log_4{6} = \frac{\ln{6}}{\ln{4}}\) \(\log_4{w} = \frac{\ln{w}}{\ln{4}}\) Now, substitute these back into the expression we obtained in Step 2: \(\log_4{6w} = \frac{\ln{6}}{\ln{4}} + \frac{\ln{w}}{\ln{4}}\)

Using the product rule and optionally the change of base formula, we've broken down the given expression \(\log_4{6w}\) into the sum of logarithms: \(\log_4{6w} = \log_4{6} + \log_4{w}\) Or, alternatively when converted to natural logarithms: \(\log_4{6w} = \frac{\ln{6}}{\ln{4}} + \frac{\ln{w}}{\ln{4}}\) Either of these can be a valid final answer depending on the requirements or context of the problem.