Logarithm Q: 4log base3 (2x+1) - 1/2log base 3 (3x-4) -....
- Thread starter currypuff
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skeeter
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Re: Logarithm Question
\(\displaystyle 4\log_3(2x+1) - \frac{1}{2} \log_3(3x-4) - \log_7(x+1)\)
start by changing the log base 7 to log base 3 ...
\(\displaystyle 4\log_3(2x+1) - \frac{1}{2} \log_3(3x-4) - \frac{\log_3(x+1)}{\log_3{7}}\)
use the power property for logs ...
\(\displaystyle \log_3(2x+1)^4 - \log_3 \sqrt{3x-4} - \log_3(x+1)^{\frac{1}{\log_3{7}}\)
combine the base 3 logs using the quotient property for logs ...
\(\displaystyle \log_3 \left(\frac{(2x+1)^4}{\sqrt{3x-4} \cdot (x+1)^{\frac{1}{\log_3{7}}}} \right)\)
\(\displaystyle 4\log_3(2x+1) - \frac{1}{2} \log_3(3x-4) - \log_7(x+1)\)
start by changing the log base 7 to log base 3 ...
\(\displaystyle 4\log_3(2x+1) - \frac{1}{2} \log_3(3x-4) - \frac{\log_3(x+1)}{\log_3{7}}\)
use the power property for logs ...
\(\displaystyle \log_3(2x+1)^4 - \log_3 \sqrt{3x-4} - \log_3(x+1)^{\frac{1}{\log_3{7}}\)
combine the base 3 logs using the quotient property for logs ...
\(\displaystyle \log_3 \left(\frac{(2x+1)^4}{\sqrt{3x-4} \cdot (x+1)^{\frac{1}{\log_3{7}}}} \right)\)