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Sagot :
h(x)=2^(x-3)
replace h(x) with y:
y=2^(x-3)=(2^x)*(2^-3)=(2^x)/(2^3)=(2^x)/8
y=2^x/8
solve for x:
8y=2^x
log2(8y)=x
replace h(x) with y:
y=2^(x-3)=(2^x)*(2^-3)=(2^x)/(2^3)=(2^x)/8
y=2^x/8
solve for x:
8y=2^x
log2(8y)=x
Answer:
[tex]f^{-1}(x)=2^x+3[/tex]
Step-by-step explanation:
Given:
[tex]y=\log_2(x-3)[/tex]
To find the inverse of the given function, make x the subject.
[tex]\textsf{Apply log law}: \quad \log_ab=c \iff a^c=b[/tex]
[tex]\implies 2^y=x-3[/tex]
Add 3 to both sides:
[tex]\implies x=2^y+3[/tex]
Swap x for [tex]f^{-1}(x)[/tex] and y for x:
[tex]\implies f^{-1}(x)=2^x+3[/tex]
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