Answer:
-9/8
Explanation:
You want a full simplification of ...
![3\cdot\log_{(4)/(9)}{\sqrt[4]{(27)/(8)}}](https://img.qammunity.org/2024/formulas/mathematics/high-school/d4rmnec70v84czdt0matrmejz91szyvdfb.png)
Rules of logarithms
The relevant rules of logarithms are ...
log(ab) = log(a) +log(b)
log(a/b) = log(a) -log(b)
log(a^b) = b·log(a)
logₐ(b) = log(b)/log(a)
Application
![3\cdot\log_{(4)/(9)}{\sqrt[4]{(27)/(8)}}=(3)/(4)\log_{(4)/(9)}\left((27)/(8)\right)=(3)/(4)\cdot(\log(27)/(8))/(\log(4)/(9))\\\\\\=(3(\log(3^3)-\log(2^3)))/(4(\log(2^2)-\log(3^2)))=(3(3\cdot\log(3)-3\cdot\log(2)))/(4(2\cdot\log(2)-2\cdot\log(3)))\\\\\\=(9(\log(3)-\log(2)))/(8(\log(2)-\log(3)))=\boxed{-(9)/(8)}](https://img.qammunity.org/2024/formulas/mathematics/high-school/nlo9fwoiik9nwzdemw75rialgml8d47553.png)
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Additional comment
We have reduced everything to the difference of the logs of 2 and 3. You could stop the reduction to smallest parts at any point where you recognize things that will cancel. You could write numerator and denominator in terms of powers of (3/2), for example.