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Function Notation with Logs and Exponentials  Problem 1
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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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Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
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Function Notation with Logs and Exponentials
Problem 1 5,747 viewsg(x) = 3[log_{3}(5x − 1)]² + 1
g(2) = 
Function Notation with Logs and Exponentials
Problem 2 5,280 viewsAmount of radioactive material decays such that the amount remaining after time, t, is given by f(t) = 100 ⋅ 2^{t} grams.
a) initial amount of materialb) amount after 4 years?c) When will there be 50 grams left? 
Function Notation with Logs and Exponentials
Problem 3 1,260 views
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