Taking inverse of a function that contains exponential term
$begingroup$
Consider we are given such a function
$$f(x) = 2^{2x-1}+3$$
How does one find $f^{-1}(x)$? Here I noticed that it is a bit different to take inverse of exponential terms. My attempt is as follows
Let $y = 2^{2x-1} +3$
$$ y - 3 = 2^{2x-1}$$
$$2y-6 = 2^{2x}$$
However, I dont think that is how exactly it works. Could you assist me with what I'm missing?
Regards
functions
asked Dec 18 '18 at 20:40
EnzoEnzo
19917
$endgroup$
|
show 2 more comments
$begingroup$
Consider we are given such a function
$$f(x) = 2^{2x-1}+3$$
How does one find $f^{-1}(x)$? Here I noticed that it is a bit different to take inverse of exponential terms. My attempt is as follows
Let $y = 2^{2x-1} +3$
$$ y - 3 = 2^{2x-1}$$
$$2y-6 = 2^{2x}$$
However, I dont think that is how exactly it works. Could you assist me with what I'm missing?
Regards
functions
asked Dec 18 '18 at 20:40
EnzoEnzo
19917
$endgroup$
2
$begingroup$
Looks good. Now take the $log()$ of both sides.
$endgroup$
– Michael
Dec 18 '18 at 20:43
$begingroup$
@Michael Are we supposed to take the log? I dont also know that.
$endgroup$
– Enzo
Dec 18 '18 at 20:43
$begingroup$
Yea, you take the log to deal with the exponent.
$endgroup$
– Don Thousand
Dec 18 '18 at 20:44
$begingroup$
@DonThousand Cannot we also just substitute $2^{2x}$ in $f(x)$?
$endgroup$
– Enzo
Dec 18 '18 at 20:45
$begingroup$
What do you mean? How would you do that?
$endgroup$
– Don Thousand
Dec 18 '18 at 20:45
|
show 2 more comments
$begingroup$
Consider we are given such a function
$$f(x) = 2^{2x-1}+3$$
How does one find $f^{-1}(x)$? Here I noticed that it is a bit different to take inverse of exponential terms. My attempt is as follows
Let $y = 2^{2x-1} +3$
$$ y - 3 = 2^{2x-1}$$
$$2y-6 = 2^{2x}$$
However, I dont think that is how exactly it works. Could you assist me with what I'm missing?
Regards
functions
asked Dec 18 '18 at 20:40
EnzoEnzo
19917
$endgroup$
Consider we are given such a function
$$f(x) = 2^{2x-1}+3$$
How does one find $f^{-1}(x)$? Here I noticed that it is a bit different to take inverse of exponential terms. My attempt is as follows
Let $y = 2^{2x-1} +3$
$$ y - 3 = 2^{2x-1}$$
$$2y-6 = 2^{2x}$$
However, I dont think that is how exactly it works. Could you assist me with what I'm missing?
Regards
functions
functions
asked Dec 18 '18 at 20:40
EnzoEnzo
19917
asked Dec 18 '18 at 20:40
EnzoEnzo
19917
asked Dec 18 '18 at 20:40
EnzoEnzo
19917
asked Dec 18 '18 at 20:40
EnzoEnzo
19917
asked Dec 18 '18 at 20:40
EnzoEnzo
19917
19917
2
$begingroup$
Looks good. Now take the $log()$ of both sides.
$endgroup$
– Michael
Dec 18 '18 at 20:43
$begingroup$
@Michael Are we supposed to take the log? I dont also know that.
$endgroup$
– Enzo
Dec 18 '18 at 20:43
$begingroup$
Yea, you take the log to deal with the exponent.
$endgroup$
– Don Thousand
Dec 18 '18 at 20:44
$begingroup$
@DonThousand Cannot we also just substitute $2^{2x}$ in $f(x)$?
$endgroup$
– Enzo
Dec 18 '18 at 20:45
$begingroup$
What do you mean? How would you do that?
$endgroup$
– Don Thousand
Dec 18 '18 at 20:45
|
show 2 more comments
2
$begingroup$
Looks good. Now take the $log()$ of both sides.
$endgroup$
– Michael
Dec 18 '18 at 20:43
$begingroup$
@Michael Are we supposed to take the log? I dont also know that.
$endgroup$
– Enzo
Dec 18 '18 at 20:43
$begingroup$
Yea, you take the log to deal with the exponent.
$endgroup$
– Don Thousand
Dec 18 '18 at 20:44
$begingroup$
@DonThousand Cannot we also just substitute $2^{2x}$ in $f(x)$?
$endgroup$
– Enzo
Dec 18 '18 at 20:45
$begingroup$
What do you mean? How would you do that?
$endgroup$
– Don Thousand
Dec 18 '18 at 20:45
2
2
$begingroup$
Looks good. Now take the $log()$ of both sides.
$endgroup$
– Michael
Dec 18 '18 at 20:43
$begingroup$
Looks good. Now take the $log()$ of both sides.
$endgroup$
– Michael
Dec 18 '18 at 20:43
$begingroup$
@Michael Are we supposed to take the log? I dont also know that.
$endgroup$
– Enzo
Dec 18 '18 at 20:43
$begingroup$
@Michael Are we supposed to take the log? I dont also know that.
$endgroup$
– Enzo
Dec 18 '18 at 20:43
$begingroup$
Yea, you take the log to deal with the exponent.
$endgroup$
– Don Thousand
Dec 18 '18 at 20:44
$begingroup$
Yea, you take the log to deal with the exponent.
$endgroup$
– Don Thousand
Dec 18 '18 at 20:44
$begingroup$
@DonThousand Cannot we also just substitute $2^{2x}$ in $f(x)$?
$endgroup$
– Enzo
Dec 18 '18 at 20:45
$begingroup$
@DonThousand Cannot we also just substitute $2^{2x}$ in $f(x)$?
$endgroup$
– Enzo
Dec 18 '18 at 20:45
$begingroup$
What do you mean? How would you do that?
$endgroup$
– Don Thousand
Dec 18 '18 at 20:45
$begingroup$
What do you mean? How would you do that?
$endgroup$
– Don Thousand
Dec 18 '18 at 20:45
|
show 2 more comments
1 Answer
1
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oldest
votes
$begingroup$
Your approach is correct. Note that $2^{2x} >0$ and thus you can always apply the $ln$ function (as mentioned-hinted in the comments). But, you also need that $2y-6 > 0 Leftrightarrow y > 3$ since you apply the $ln$ to that side as well of course. Taking this constrict into consideration, it is :
$$ln(2y-6) = ln 2^{2x} Leftrightarrow ln(2y-6) = 2x ln 2 Leftrightarrow x = frac{ln(2y-6)}{ln 4}$$
Now, interchange $x leftrightarrow y$ and yield :
$$y = frac{ln(2x-6)}{ln 4} implies f^{-1}(x) = frac{ln(2x-6)}{ln 4}$$
Don't forget that it must be $2x-6 >0 Leftrightarrow x > 3$ and $y >3 implies f(x) >3$ from the initial constrict.
answered Dec 18 '18 at 20:49
RebellosRebellos
14.8k31248
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add a comment |
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1 Answer
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$begingroup$
Your approach is correct. Note that $2^{2x} >0$ and thus you can always apply the $ln$ function (as mentioned-hinted in the comments). But, you also need that $2y-6 > 0 Leftrightarrow y > 3$ since you apply the $ln$ to that side as well of course. Taking this constrict into consideration, it is :
$$ln(2y-6) = ln 2^{2x} Leftrightarrow ln(2y-6) = 2x ln 2 Leftrightarrow x = frac{ln(2y-6)}{ln 4}$$
Now, interchange $x leftrightarrow y$ and yield :
$$y = frac{ln(2x-6)}{ln 4} implies f^{-1}(x) = frac{ln(2x-6)}{ln 4}$$
Don't forget that it must be $2x-6 >0 Leftrightarrow x > 3$ and $y >3 implies f(x) >3$ from the initial constrict.
answered Dec 18 '18 at 20:49
RebellosRebellos
14.8k31248
$endgroup$
add a comment |
$begingroup$
Your approach is correct. Note that $2^{2x} >0$ and thus you can always apply the $ln$ function (as mentioned-hinted in the comments). But, you also need that $2y-6 > 0 Leftrightarrow y > 3$ since you apply the $ln$ to that side as well of course. Taking this constrict into consideration, it is :
$$ln(2y-6) = ln 2^{2x} Leftrightarrow ln(2y-6) = 2x ln 2 Leftrightarrow x = frac{ln(2y-6)}{ln 4}$$
Now, interchange $x leftrightarrow y$ and yield :
$$y = frac{ln(2x-6)}{ln 4} implies f^{-1}(x) = frac{ln(2x-6)}{ln 4}$$
Don't forget that it must be $2x-6 >0 Leftrightarrow x > 3$ and $y >3 implies f(x) >3$ from the initial constrict.
answered Dec 18 '18 at 20:49
RebellosRebellos
14.8k31248
$endgroup$
add a comment |
$begingroup$
Your approach is correct. Note that $2^{2x} >0$ and thus you can always apply the $ln$ function (as mentioned-hinted in the comments). But, you also need that $2y-6 > 0 Leftrightarrow y > 3$ since you apply the $ln$ to that side as well of course. Taking this constrict into consideration, it is :
$$ln(2y-6) = ln 2^{2x} Leftrightarrow ln(2y-6) = 2x ln 2 Leftrightarrow x = frac{ln(2y-6)}{ln 4}$$
Now, interchange $x leftrightarrow y$ and yield :
$$y = frac{ln(2x-6)}{ln 4} implies f^{-1}(x) = frac{ln(2x-6)}{ln 4}$$
Don't forget that it must be $2x-6 >0 Leftrightarrow x > 3$ and $y >3 implies f(x) >3$ from the initial constrict.
answered Dec 18 '18 at 20:49
RebellosRebellos
14.8k31248
$endgroup$
Your approach is correct. Note that $2^{2x} >0$ and thus you can always apply the $ln$ function (as mentioned-hinted in the comments). But, you also need that $2y-6 > 0 Leftrightarrow y > 3$ since you apply the $ln$ to that side as well of course. Taking this constrict into consideration, it is :
$$ln(2y-6) = ln 2^{2x} Leftrightarrow ln(2y-6) = 2x ln 2 Leftrightarrow x = frac{ln(2y-6)}{ln 4}$$
Now, interchange $x leftrightarrow y$ and yield :
$$y = frac{ln(2x-6)}{ln 4} implies f^{-1}(x) = frac{ln(2x-6)}{ln 4}$$
Don't forget that it must be $2x-6 >0 Leftrightarrow x > 3$ and $y >3 implies f(x) >3$ from the initial constrict.
answered Dec 18 '18 at 20:49
RebellosRebellos
14.8k31248
answered Dec 18 '18 at 20:49
RebellosRebellos
14.8k31248
answered Dec 18 '18 at 20:49
RebellosRebellos
14.8k31248
answered Dec 18 '18 at 20:49
RebellosRebellos
14.8k31248
14.8k31248
add a comment |
add a comment |
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2
$begingroup$
Looks good. Now take the $log()$ of both sides.
$endgroup$
– Michael
Dec 18 '18 at 20:43
$begingroup$
@Michael Are we supposed to take the log? I dont also know that.
$endgroup$
– Enzo
Dec 18 '18 at 20:43
$begingroup$
Yea, you take the log to deal with the exponent.
$endgroup$
– Don Thousand
Dec 18 '18 at 20:44
$begingroup$
@DonThousand Cannot we also just substitute $2^{2x}$ in $f(x)$?
$endgroup$
– Enzo
Dec 18 '18 at 20:45
$begingroup$
What do you mean? How would you do that?
$endgroup$
– Don Thousand
Dec 18 '18 at 20:45