Answer:
[tex]4\log_{\frac{1}{2}}w+(2\log_{\frac{1}{2}}u-3\log_{\frac{1}{2}}v)=\log_{\frac{1}{2}}(\frac{w^4u^2}{v^3})[/tex]
Step-by-step explanation:
Given : Expression [tex]4\log_{\frac{1}{2}}w+(2\log_{\frac{1}{2}}u-3\log_{\frac{1}{2}}v)[/tex]
To write : As a single logarithm?
Solution :
[tex]4\log_{\frac{1}{2}}w+(2\log_{\frac{1}{2}}u-3\log_{\frac{1}{2}}v)[/tex]
Remove parenthesis,
[tex]=4\log_{\frac{1}{2}}w+2\log_{\frac{1}{2}}u-3\log_{\frac{1}{2}}v[/tex]
Simplify each term by applying logarithmic property, [tex]a\log x=\log x^a[/tex]
[tex]=\log_{\frac{1}{2}}w^4+\log_{\frac{1}{2}}u^2-\log_{\frac{1}{2}}v^3[/tex]
Use the product property of logarithms, [tex]\log_bx+\log_b y=\log_b (xy)[/tex]
[tex]=\log_{\frac{1}{2}}w^4u^2-\log_{\frac{1}{2}}v^3[/tex]
Use the quotient property of logarithms, [tex]\log_bx-\log_b y=\log_b (\frac{x}{y})[/tex]
[tex]=\log_{\frac{1}{2}}(\frac{w^4u^2}{v^3})[/tex]
Therefore,
[tex]4\log_{\frac{1}{2}}w+(2\log_{\frac{1}{2}}u-3\log_{\frac{1}{2}}v)=\log_{\frac{1}{2}}(\frac{w^4u^2}{v^3})[/tex]
