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Question: How do you condense \(3\ln 3 + \ln 9\)?...

How do you condense 3ln⁡3+ln⁡93\ln 3 + \ln 9?

Explanation

Solution

In order to determine the condensed or we can say simplified form of the above question, rewrite the expression using the property of logarithm nlog⁡m=log⁡mnn\log m = \log {m^n}by taking n=3 n = 3\,and m=3m = 3 then combine both of the logarithmic values into single logarithm by using the identity of addition of logarithm which says log⁡b(m)+log⁡b(n)=log⁡b(mn){\log _b}(m) + {\log _b}(n) = {\log _b}(mn).By solving this you’ll get your desired condensed form.

Complete step by step solution:
We are Given an expression 3ln⁡3+ln⁡93\ln 3 + \ln 9 So to condense or simplify the expression we’ll be using some of the properties of logarithm.

=3ln⁡3+ln⁡9 = 3\ln 3 + \ln 9
3ln⁡33\ln 3can be written as ln⁡33\ln {3^3}using the property of logarithm nlog⁡m=log⁡mnn\log m = \log {m^n}where n=3 n = 3\,and m=3m = 3.

Replacing 3ln⁡33\ln 3with in the expression ln⁡33\ln {3^3}
=ln⁡33+ln⁡9= \ln {3^3} + \ln 9

As we know that any addition of two logarithmic values can be expressed aslog⁡b(m)+log⁡b(n)=log⁡b(mn){\log _b}(m) + {\log_b}(n) = {\log _b}(mn)here, m and n are 33{3^3}and 99respectively

Now our equation becomes
=ln⁡(33×9) =ln⁡(27×9) =ln⁡(243)  = \ln \left( {{3^3} \times 9} \right) \\\ = \ln \left( {27 \times 9} \right) \\\ = \ln \left( {243} \right) \\\
Therefore, the condensed form of the expression 3ln⁡3+ln⁡93\ln 3 + \ln 9is equal to ln⁡(243)\ln \left( {243} \right).

Additional Information:
1.Value of constant ‘e’ is equal to 2.718282.71828.

2.A logarithm is basically the reverse of a power or we can say when we calculate a logarithm of any number, we actually undo an exponentiation.

3.Any multiplication inside the logarithm can be transformed into addition of two separate logarithm values.
log⁡b(mn)=log⁡b(m)+log⁡b(n){\log _b}(mn) = {\log _b}(m) + {\log _b}(n)

4. Any division inside the logarithm can be transformed into subtraction of two separate logarithm values.
log⁡b(mn)=log⁡b(m)−log⁡b(n){\log _b}\left( {\dfrac{m}{n}} \right) = {\log _b}(m) - {\log _b}(n)

5. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
nlog⁡m=log⁡mnn\log m = \log {m^n}

Note: 1.Don’t forget to cross-check your answer at least once.
2.ln⁡\ln is known as the “natural log” which is having base ee.