Logarithm question for algebra 2

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Logarithm question for algebra 2

Mellie Apr 26, 2018

TheXSquaredFactor · Answer 1 · 2018-04-26T10:32:43

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