I'm working with a student in Algebra II. There are a couple of problems in the chapter on log functions that I can't figure out, nor can I find the solutions in the chapter (I sound like a student, myself). Here they are: Write the inverse function for this 1. y = log 2x 2 The answer given is y = 2^x-1 2. y = log(x+1) No answer given Please show me how you get the answers. Thank you.
1. I'm assuming the original equation is y = (log base 2 of (x)) + 1. f-1(x) is the inverse function. log base 2 of x - 1 Remember: f(f-1(x)) = x f(f-1(x)) = log base 2 of (f-1(x)) + 1 log base 2 of (f-1(x)) = x - 1 f-1(x) = 2^(x-1) ================== f(x) = log(x+1) log(f-1(x) + 1) = x f-1(x) + 1 = 10^x f-1(x) = 10^x - 1
OK, here's my response: First, change each log equation into an exponential equation: - keep the base the same. - since a log = an exponent, whatever is on the OTHER side of the equal sign from the word "log" is the exponent. - the other term is what it equals. Second, to find the inverse of a function, switch the x and the y. Solve for y if necessary. OK, problem #1. To get the answer you list, the problem I'm starting with is: y = log (base 2) of (x+1) First, switch the problem to exponential form: 2 to the y = (x+1) Now, find the inverse: switch the x and y: 2 to the (x) = y+1. (2 to the x ) -1 = y If that wasn't the original equation, let me know. Number 2: y = log (base 10)-- that's assumed if there's no other value listed) of (x+1) Rewrite in exponential form: 10 to the y = (x+1) Now switch the x and y: 10 to the x = (y+1) Subtract 1 from both sides to solve for y: 10 to the x - 1 = y.
What I say is, "Wow", and, "Thanks". I understood yours well, Alice. Yes, you both had the equation right. I forgot it wouldn't line up on here. Mike, the part I don't understand in your solution is : Remember: f(f-1(x)) = x I sure didn't remember that. Would you explain it to me (a former middle school math teacher)? Oh, and I really appreciated the photo. Thank you both very much.
It's just a property of the inverse function. For example, f(x) = 4x. Therefore f-1(x) = x/4. f( f-1(x) ) = f(x/4) = (4x)/4) = x. Another example, f(x) = sin(x). Therefore f-1(x) = arcsin(x). f( f-1(x) ) = f(arcsin(x)) = sin(arcsin(x)) = x.
If you have f(x)=x+2, you know that f-1(x)=x-2 f(1)=3 f-1(3)=1 So, f(f-1(x))=x because f(f-1(3))=f(1)=3
Since superscripts are impractical on these forums, how about adopting the old convention, from line-printer days, of using a caret to indicate that what follows it is an exponent? Then "x raised to the second power" as x^2, and Mike's f - 1(x) is f^-1(x).
You're welcome, 'Daisy! As to the explanation: f( f^-1(x))=x: Finding the inverse means switching the x and y. Remember, f^-1(x) means the inverse of x, or y. So, to put it into words, f( f^-1(x))=x: means take f(the inverse of x) or Plug the inverse of x into the original function in place of x. As long as f(x) and (f^-1)(x) are inverses, you'll end up with x. For what it's worth, this is NOT the explanation I would give to a high school student who is struggling!
Nor to her tutor, who is still struggling. My student won't be required to know this, I'm pretty sure. If I didn't think I were helping her, I would bow out, but I know it has been a benefit to her. However, I'm sure not a candidate to teach an algebra II class.
