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Use the Product Rule of Logarithms to write the completely expanded expression equivalent to [tex]\ln (6a \cdot 9b)[/tex]. Make sure to use parentheses around your logarithm functions [tex]\ln (x + y)[/tex].

Note: When entering natural log in your answer, enter lowercase "ln" as "In". There is no "natural log" button on the Alta keyboard.

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Sagot :

GinnyAnsweridn
Sure! Let's solve the problem step-by-step using the Product Rule of Logarithms.

The Product Rule of Logarithms states that for any positive numbers [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:

[tex]\[ \ln(x \cdot y) = \ln(x) + \ln(y) \][/tex]

We need to apply this rule to the expression [tex]\( \ln(6a + 9b) \)[/tex].

First, let's factor the expression inside the logarithm:

[tex]\[ 6a + 9b \][/tex]

Notice that we can factor out a 3 from both terms:

[tex]\[ 6a + 9b = 3(2a + 3b) \][/tex]

Now, we have:

[tex]\[ \ln(6a + 9b) = \ln(3(2a + 3b)) \][/tex]

According to the Product Rule of Logarithms, we can separate the logarithm of the product into the sum of logarithms:

[tex]\[ \ln(3(2a + 3b)) = \ln(3) + \ln(2a + 3b) \][/tex]

So, the completely expanded expression equivalent to [tex]\( \ln(6a + 9b) \)[/tex] is:

[tex]\[ \boxed{\text{In}(6a + 9b) = \text{In}(3) + \text{In}(2a + 3b)} \][/tex]
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