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Function Notation with Logs and Exponentials - Problem 1
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Solving a logarithmic equation in a function notation. So whenever we see something in function notation what we have to do is just plug in the number in the parenthesis. With log there’s no difference. So behind me I have a fairly elaborate log equation and we are asked to find g of 2.

Basically what we want to do is plug 2 in wherever we see x. There’s only one x in this equation so we plug in 2 there, 5 times 2 minus 1² plus 1. So now using order of operations, we know we need to simplify our parenthesis first. 5 times 2 is 10 minus 9 is sorry minus 1 is 9, getting a little bit ahead of myself, so what we have right here is log base 3 of 9² to 3 and the plus 1.

So hopefully you are getting into a point where you can start to recognize what these log things actually do. And this is basically saying 3 to what power is equal to 9? 3² is equal to 9 so this term right here is just 2.

From here we know that 3 times 2² plus 1, we just have a simple equation solve it out. 2² is 4, 3 times 2² 12 plus 1, 13. So just by using our function notation and our properties of logs, we were able to solve this out.

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