Discover a world of knowledge and community-driven answers at IDNLearn.com today. Discover comprehensive answers from knowledgeable members of our community, covering a wide range of topics to meet all your informational needs.
Sagot :
To determine the equation of the inverse of the function [tex]\( y = 4^{2x + 9} \)[/tex], we need to follow a series of steps to solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex] and then interchange [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
1. Rewrite the given function:
[tex]\[ y = 4^{2x + 9} \][/tex]
2. Switch [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to find the inverse function:
[tex]\[ x = 4^{2y + 9} \][/tex]
3. Take the logarithm base 4 of both sides to solve for [tex]\( y \)[/tex]:
Recall that if [tex]\( a = b^c \)[/tex], then [tex]\( \log_b(a) = c \)[/tex].
[tex]\[ \log_4(x) = 2y + 9 \][/tex]
4. Isolate [tex]\( y \)[/tex]:
[tex]\[ \log_4(x) - 9 = 2y \][/tex]
[tex]\[ y = \frac{\log_4(x) - 9}{2} \][/tex]
Thus, the inverse function of [tex]\( y = 4^{2x + 9} \)[/tex] is:
[tex]\[ y = \frac{\log_4(x) - 9}{2} \][/tex]
Therefore, the correct answer is:
[tex]\[ y = \frac{\log_4(x) - 9}{2} \][/tex]
1. Rewrite the given function:
[tex]\[ y = 4^{2x + 9} \][/tex]
2. Switch [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to find the inverse function:
[tex]\[ x = 4^{2y + 9} \][/tex]
3. Take the logarithm base 4 of both sides to solve for [tex]\( y \)[/tex]:
Recall that if [tex]\( a = b^c \)[/tex], then [tex]\( \log_b(a) = c \)[/tex].
[tex]\[ \log_4(x) = 2y + 9 \][/tex]
4. Isolate [tex]\( y \)[/tex]:
[tex]\[ \log_4(x) - 9 = 2y \][/tex]
[tex]\[ y = \frac{\log_4(x) - 9}{2} \][/tex]
Thus, the inverse function of [tex]\( y = 4^{2x + 9} \)[/tex] is:
[tex]\[ y = \frac{\log_4(x) - 9}{2} \][/tex]
Therefore, the correct answer is:
[tex]\[ y = \frac{\log_4(x) - 9}{2} \][/tex]
Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. IDNLearn.com is your go-to source for accurate answers. Thanks for stopping by, and come back for more helpful information.