We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

#1**+3 **

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. |

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

TheXSquaredFactor Apr 26, 2018