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Logarithm question for algebra 2

 Apr 26, 2018
 #1
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I think I will complete a and c. I think you can manage the rest on your own. You can use my work as a model. 

 

a) 

 

\(10^{x^2}=320\) Take the logarithm base 10 of both sides, which is exactly that the directions say!
\(\log_{10}\left(10^{x^2}\right)=\log_{10}(320)\) Because of the properties of logarithms, the exponent can be extracted, and it will become the coefficient of the logarithm.
\(x^2\log_{10}(10)=\log_{10}(320)\) The \(\log_{10}(10)\) simiplifies to 1 because the base and the argument are the same value. 
\(x^2=\log_{10}(320)\) Take the square root of both sides. 
\(|x|=\sqrt{\log_{10}(320)}\) Of course, the absolute value creates two solutions: the positive and negative one.
\(x=\pm\sqrt{\log_{10}(320)}\)  
   

 

c)

 

\(5^{2x}=200\) Take the logarithm base 10 of both sides again.
\(\log_{10}(5^{2x})=\log_{10}(200)\) Yet again, the exponent becomes the coefficient. This is a basic property of logarithms.
\(2x\log_{10}(5)=\log_{10}(200)\) Divide by \(2\log_{10}(5)\) to isolate x. 
\(x=\frac{\log_{10}(200)}{2\log_{10}(5)}\)  
   
 Apr 26, 2018

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