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Find the inverse of the function

y = log₂ (x-3)

Sagot :

MapleCool12
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
semsee45

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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