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. |
\(x=\frac{\log_{10}(200)}{2\log_{10}(5)}\) | |