Discussion in 'Secondary Education' started by Upsadaisy, Apr 18, 2012.

Joined:
Aug 2, 2002
Messages:
18,938
681

Apr 18, 2012

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)

2. 3. ### MikeTeachesMathDevotee

Joined:
Nov 5, 2011
Messages:
1,163
3

Apr 18, 2012

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 4. ### AliceaccMultitudinous

Joined:
Apr 12, 2006
Messages:
27,534
6

Apr 18, 2012

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.

5. ### MikeTeachesMathDevotee

Joined:
Nov 5, 2011
Messages:
1,163
3

Apr 18, 2012

^ What she said :lol:

Joined:
Aug 2, 2002
Messages:
18,938
681

Apr 18, 2012

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.

7. ### MikeTeachesMathDevotee

Joined:
Nov 5, 2011
Messages:
1,163
3

Apr 18, 2012

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.

8. ### orangeteaConnoisseur

Joined:
Jan 12, 2012
Messages:
1,600
0

Apr 18, 2012

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

9. ### orangeteaConnoisseur

Joined:
Jan 12, 2012
Messages:
1,600
0

Apr 18, 2012

Oh, and it goes both ways. If f(x) and g(x) are inverses, then...

f(g(x)) = x and g(f(x)) = x

10. ### TeacherGroupieModerator

Joined:
May 13, 2005
Messages:
29,807
1,170

Apr 18, 2012

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).

11. ### AliceaccMultitudinous

Joined:
Apr 12, 2006
Messages:
27,534
6

Apr 18, 2012

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!

Joined:
Aug 2, 2002
Messages:
18,938
681

Apr 18, 2012

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.