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# Logarithm question for algebra 2

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Logarithm question for algebra 2 Apr 26, 2018

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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