Question

A problem dealing with Logarithms.?

Answer 1

You can use these rules here:

1.\[ a \cdot \log_3x=\log_3x^a\]

2. \[\log_3a+\log_3b=\log_3ab\]
3.\[\log_3a-\log_3b=\log_3\frac{a}{b}\]

Try it, please?

Answer 2

Let me do the first step (using Rule #1):\[\frac{ 1 }{ 2 }\cdot \log_3x=\log_3x^{\frac{1}{2}}\]

Can you do the next one?

Answer 3

@tinaballerina: would you do that for me?