Finding the inverse of ##y=2^{x}##

[chwala]

Homework Statement

find the inverse of the function $y=2^{x}$
ok i know the steps but why is this question awarded 1 mark...

Homework Equations

3. The Attempt at a Solution [/B]
$y= 2^{x}$
→$ln y= xln 2$
→$x= ln y/ln 2$

 

[neilparker62]

All good - can you rewrite in perhaps more elegant form ?

 

[symbolipoint]

EDIT: I am not sure if I should delete this post or not. See the post #5. That one is better but the notation is not.

You can use the quick way or you can use the more formally correct way. Logarithm function IS inverse of exponential function.

The quick way:

2^y=x, just switching x and y roles.

log(2^y)=log(x)
y*log(2)=log(x)
y=log(x)/log(2)

This could also be shown as y=log(x^(1/log(2)))

 

[neilparker62]

The quick way: ...

All good again - but still looking for the elegant way ...!

 

[symbolipoint]

I am no longer comfortable with post #3.

Say, h(x) is the inverse of y=2^x.

2^(h(x))=x
putting this into exponential logarithmic form,
log_2_of_(x)=h(x)
h(x)=log_2_of(x)

I am unsure how to handle the notation better. Base is 2.

 

[neilparker62]

Perfect: inverse function of y = 2^x is y = log₂x. And to answer the OP's question, it's only one mark because it rests on the basic definition of a log function:

$$ y = a^x ⇔ log_{a}y=x $$

 

[Delta2]

To perfectly and rigorously define the inverse function we need not only to find the formula of the function but the domain and range sets as well...

 

[chwala]

is t

Perfect: inverse function of y = 2^x is y = log₂x. And to answer the OP's question, it's only one mark because it rests on the basic definition of a log function:

$$ y = a^x ⇔ log_{a}y=x $$

is that all to it?

 

[member 587159]

Perfect: inverse function of y = 2^x is y = log₂x. And to answer the OP's question, it's only one mark because it rests on the basic definition of a log function:

$$ y = a^x ⇔ log_{a}y=x $$

Just writing this down doesn't make sense.Here is work to do (is log well defined? etc.)

 

[SammyS]

Homework Statement

find the inverse of the function $y=2^{x}$
ok i know the steps but why is this question awarded 1 mark...

Homework Equations

The Attempt at a Solution

$y= 2^{x}$
→$ln y= xln 2$
→$x= ln y/ln 2$

Wow ! In a time interval of 31 minutes, you had 6 posts in this thread replying to OP or to the other posts.

You ask, "why is this question awarded 1 mark" . (I suppose you mean: "Why was this solution awarded only 1 mark ?") I'm guessing that '1 mark' indicates that it's a rather unsatisfactory solution. Not being the grader, nor the person setting up the marking scheme, we can only speculate regarding the marking here. I presume the marking rubric depends upon details of the material that was presented to the student. We are left to speculate in our responses.

If the question had been:

Solve $\ y= 2^{x}\$ for $\ x\$,​

then the given solution should be perfectly satisfactory.

In a course which formally deals with functions and their inverses, the following problem statement might be less ambiguous.

Given that $\displaystyle \ f(x) = 2^x\$, what is the inverse function, $\ \displaystyle \ f^{-1}(x) \,?$​

My expected response to this is: $\ \displaystyle \ f^{-1}(x) = \log_2 (x) \,.$
It's possible that course material is presented so that $\ y\$ is generally a function of $\ x \,.\$ In this case the expected response likely is: $\ \displaystyle \ y = \log_2 (x) \,.$ (It's easy to see possible confusion with this.)

Another presentation of this material might elicit the following as the preferred response.

$\ \displaystyle \ x = \log_2 (y) \,.$​

This is often a direct application of the definition of the logarithm function given in a basic algebra course. (in USA)

The best advice I can give is to ask the person with responsibility for the marking.

Added in Edit:
Of course, domain and range are important in defining any function, as pointed out by Δ².

 

[chwala]

Sammy i think your approach is sufficient, taking logs on both sides to base 2 and hence getting the inverse function.

 

[FactChecker]

Did you restrict the legitimate values of y? Forgetting that probably deserves a deduction of some type.

 

[chwala]

i think i wrote the problem incorrectly, it was meant to be a proof of $ff^-1(x)=x$ problem,... where $f(x)=2^x$ my apologies...

 

[SammyS]

i think i wrote the problem incorrectly, it was meant to be a proof of $ff^-1(x)=x$ problem,... where $f(x)=2^x$ my apologies...

To include the 1 in the superscript, place -1 in braces. Also, an additional set of parentheses will enhance readability.

You would get: $f(f^{-1}(x))=x$

from the LaTeX code: ##f(f^{-1}(x))=x## .

Regarding the problem posed in the OP:

So after finding that $f^{-1}(x) = \log _2(x)$

( or equivalently: $f^{-1}(x) = \dfrac {\ln(x)}{\ln(2)} \ \$ )​

I suppose that you were supposed to prove that $\displaystyle 2^ {\log _2(x)} = x$ .
.

 

[chwala]

To include the 1 in the superscript, place -1 in braces. Also, an additional set of parentheses will enhance readability.

You would get: $f(f^{-1}(x))=x$

from the LaTeX code: ##f(f^{-1}(x))=x## .

Regarding the problem posed in the OP:

So after finding that $f^{-1}(x) = \log _2(x)$

( or equivalently: $f^{-1}(x) = \dfrac {\ln(x)}{\ln(2)} \ \$ )​

I suppose that you were supposed to prove that $\displaystyle 2^ {\log _2(x)} = x$ .
.

right, thanks Sammy, nice to hear from you brother...

 

[chwala]

can we prove this way?
let $log_{2}x = m$
it follows that $2^m=x$
$m log 2 = log x$
$m log_{2} 2 = log_{2}x$
$m = log_{2} x$
$log_{2}x = log_{2} x$

 

[nuuskur]

What have you proved, exactly? You showed $\log _2 x = m$ implies $\log _2 x = m$, which I think is quite obvious without any explicit proof.

The map $y = 2^x$ has domain $\mathbb R$ and codomain $(0,\infty)$. It is invertible in the domain, hence its inverse is $y^{-1} = \log _2x$, where $x >0$.

Alternatively, the same exercise could be specified such that $y=2^x$, where $x\geq 2$, say. It is also invertible, but what is its inverse? Suffices to say, it has a different inverse than in the previous case. It really depends on whether this is high school mathematics or entry level calculus in university.

 

[chwala]

What have you proved, exactly? You showed $\log _2 x = m$ implies $\log _2 x = m$, which I think is quite obvious without any explicit proof.

The map $y = 2^x$ has domain $\mathbb R$ and codomain $(0,\infty)$. It is invertible in the domain, hence its inverse is $y^{-1} = \log _2x$, where $x >0$.

Alternatively, the same exercise could be specified such that $y=2^x$, where $x\geq 2$, say. It is also invertible, but what is its inverse? Suffices to say, it has a different inverse than in the previous case. It really depends on whether this is high school mathematics or entry level calculus in university.

I get your point, this is high school math. What do you mean by saying that it may have a different inverse?

 

[nuuskur]

To illustrate my point. Consider the functions $f(x) = x^2,\ x\in\mathbb R$ and $g(x) = x^2,\ x >0$. They are different, because their domains are different. One of them is invertible and the other one isn't, although, I don't think invertibility is explored in high school.

Two functions are equal if and only if their domain and codomain coincide and for every $x$ in the domain the equality $f(x) = g(x)$ holds.

I will assume all the functions in the exercises you are given are invertible. Take $y=2^x$. Identify the domain and codomain: $X=\mathbb R$ and $Y=(0,\infty)$. Then switch $x,y$ and express $y$ as a function of $x$ and get $y = \log _2x$. This is the inverse with domain $(0,\infty)$ and codomain $\mathbb R$.

In the other example $y=2^x,\ x\geq 2$. The domain is $X=[2,\infty)$ and the codomain is $Y = [4,\infty)$. The inverse is now $y=\log _2x$ with domain $[4,\infty)$ and codomain $[2,\infty)$.

Notice that domain and codomain switch places.

In high school there is perhaps talk of "range of a function", which is not the same as the codomain. Range is contained in the codomain, but not always vica versa (invertibility requires they be equal). We could define $y=2^x$ with domain and codomain $\mathbb R$, which would then not be invertible. For all intents and purposes, you may think of range as the codomain in your exercises.

 

[chwala]

To illustrate my point. Consider the functions $f(x) = x^2,\ x\in\mathbb R$ and $g(x) = x^2,\ x >0$. They are different, because their domains are different. One of them is invertible and the other one isn't, although, I don't think invertibility is explored in high school.

Two functions are equal if and only if their domain and codomain coincide and for every $x$ in the domain the equality $f(x) = g(x)$ holds.

I will assume all the functions in the exercises you are given are invertible. Take $y=2^x$. Identify the domain and codomain: $X=\mathbb R$ and $Y=(0,\infty)$. Then switch $x,y$ and express $y$ as a function of $x$ and get $y = \log _2x$. This is the inverse with domain $(0,\infty)$ and codomain $\mathbb R$.

In the other example $y=2^x,\ x\geq 2$. The domain is $X=[2,\infty)$ and the codomain is $Y = [4,\infty)$. The inverse is now $y=\log _2x$ with domain $[4,\infty)$ and codomain $[2,\infty)$.

Notice that domain and codomain switch places.

In high school there is perhaps talk of "range of a function", which is not the same as the codomain. Range is contained in the codomain, but not always vica versa (invertibility requires they be equal). We could define $y=2^x$ with domain and codomain $\mathbb R$, which would then not be invertible. For all intents and purposes, you may think of range as the codomain in your exercises.

yeah i know that...domain and range, maybe i didn't understand your 'language' thanks though..