Question

How do you find the inverse of `y = (log_2 x) +2`?

Answer 1

This process reflects the equation about `y=x`

`y=10^(log_10(2)(x-2)`

Explanation:

Using: `log_2(x) -> log_10(x)/log_10(2)`

`" "y=log_10(x)/log_10(2)+2`

`color(brown)("Subtract 2 from both sides")`

`" "y-2=log_10(x)/log_10(2)`

`color(brown)("Multiply both sides by "log_10(2))`
`color(white)(2/2)`

`" "log_10(2)(y-2)=log_10(x)`
`" "color(brown)(|ul(" ")|)`
`" "color(brown)(darr)`
`color(brown)(" Let "log_10(2)(y-2)" be "z)`

`log_10(x)=z " "vec("another way of writing this is")" " x=10^z`
`color(white)(2/2)`
`color(brown)("So by substitution for "z" we have:")`
`color(white)(2/2)`
`" "=>x=10^(log_10(2)(y-2)`

`color(brown)("Swap the letters "x" and "y" round and we have:")`
`color(white)(2/2)`

`" "color(blue)(y=10^(log_10(2)(x-2))`
`color(white)(2/2)`