"With $new_min = 5, it would give a range of 0..95 rather than 5..100 if you don't include the offset."

So the following is what is meant here?

my $x = (($tap - $min) * ($new_max - $new_min + 5) / ($max - $min));
#                                             ^^^
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No.

my $x = (($tap - $min) * ($new_max - $new_min) / ($max - $min)) + $new
+_min;
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The offset on the new range has to be outside the multiply.

stevieb: I don't know how rusty your algebra is, but it's always good to go back to first principles: what you're doing is a liner transformation from an input number to an output number. So let's derive it using everybody's favorite line, y = mx + b. The y are your output points (y0 = $new_min and y1 = $new_max), the x are you input points (x0 = $min and x1 = $max). Then let's calculate a y from a given x (where x = $tap). (We'll ignore the fact that you called the output $x, just to confuse us, until the very end.)

    #I#     y0 = m*x0 + b
    #II#    y1 = m*x1 + b

Want to eliminate b, so we're down to one unknown:

    #III#   (y1-y0) = m*(x1-x0) + (b-b)     #II#-#I#
    #IV#    (y1-y0) = m*(x1-x0)             ## simplify
    #V#     m = (y1-y0)/(x1-x0)             ## solve for m

Now we need to solve for b

    #VI#    y0 = (y1-y0)/(x1-x0)*x0 + b     #V# into #I#
    #VII#   b = y0 - x0*(y1-y0)/(x1-x0)     ## solve for b

Plug into the generic equation

    #VIII#  y = m*x + b                                         ## generic
    #IX#    y = (y1-y0)/(x1-x0)*x + y0 - x0*(y1-y0)/(x1-x0)     #V# and #VII# into #VIII#
    #X#     y = ( (y1-y0)*x - (y1-y0)*x0 )/(x1-x0) + y0         ## bring most over common denominator
    #XI#    y = ( y1*(x-x0) - y0*(x-x0) )/(x1-x0) + y0          ## group the y1's and the x1's
    #XII#   y = ( (y1-y0)*(x-x0) )/(x1-x0) + y0                 ## combine the multipliers of (x-x0)
    #XIII#  y = (x-x0)*(y1-y0)/(x1-x0) + y0                     ## reorder, remove extra parens

Convert to your notation
    x0 => $min
    x1 => $max
    x  => $tap
    y0 => $new_min
    y1 => $new_max
    y => $x

    y = (x-$min)*(y1-y0)/(x1-$min) + y0                         ## substitute for x0 into #XIII#
    y = (x-$min)*(y1-y0)/($max-$min) + y0                       ## substitute for x1 into prev
    y = ($tap-$min)*(y1-y0)/($max-$min) + y0                    ## substitute for x  into prev
    y = ($tap-$min)*(y1-$new_min)/($max-$min) + $new_min        ## substitute for y0 into prev
    y = ($tap-$min)*($new_max-$new_min)/($max-$min) + $new_min  ## substitute for y1 into prev
    $x = ($tap-$min)*($new_max-$new_min)/($max-$min) + $new_min ## substitute for y  into prev

And with my and the end semicolon, you now have the equation that typbalt89 gave: my $x = ($tap-$min)*($new_max-$new_min)/($max-$min) + $new_min;