This one does my head in. I do it fairly regularly, lots of people do, but I find everytime it comes to interpreting the results I have to slow it right down and go step by step.

**Answer:** When you log the outcome, then on the original scale, for all else constant, Y will be exp(b1*d) higher. Or,

**Y, your outcome on the original scale, will change by (exp(beta*d)-1)*100 % for a d unit change in x1.**

Where b1 is the coefficient for the variable of interest and d is the difference in that variable you are interested in.

**Here’s how we got there:**

We have

Y = a + b1x1 + b2x2 + e

right, simple. Then with logging the outcome we have,

log(Y) = a + b1x1 + b2x2 + e

right, our model is still linear.

What how do we interpret the b1 coefficient on the original scale of Y? Take the exponential of both sides (assuming we used a natural log). Then we have,

Y = exp(a + b1x1 + b2x2 + e)

which with our log/exponent rules is the equivalent of

Y = exp(a)*exp(b1x1)*exp(b2x2)*exp(e)

if we hone on variable 1, of interest, then we fix the levels of the other variables (‘for all else constant’). So we reduce to,

Y = c*exp(b1x1)

where c is some arbitrary constant.

For an increase in x1 of 10 say, we have

Y = c*exp(b1(x1+10))

which is

Y = c*exp(b1x1+b1*10)

right, and as before this becomes

Y = c*exp(b1x1)*exp(b1*10)

Now we’re getting some! So x1 compared to x1+10 (10 units higher) is difference between c*exp(b1x1) and c*exp(b1x1)*exp(b1*10). So what is the difference? **Well the latter is a multiple of exp(b1*10) higher than the former!** It goes without saying that that holds regardless of what difference you are looking at, be it 10 or 20 or 100.

So that leaves us where? With still a bit more to go. b1 is constant and fixed right, and for the sake of a comparison, so is x1, so let’s roll that term in with our constant, and lets not look at 10 but look at d for difference.

For our base,

Y = c’

where c’ is c*exp(b1x1), and our change is to

Y = c’*exp(b1*d)

Regardless of our choice of x1, if it changes by d (our difference) the outcome on the original scale will be exp(b1*d) higher. Done.

**An example with numbers:**

In this example our c (constant of everything else rolled up) is 8 (arbitrarily chosen). The last part of that shows that y2 is 7% (100*0.070) higher than y1.

And there lies the key!

**Y1, your outcome on the original scale, will change by (exp(beta*d)-1)*100 % for a d unit change in x1.**

When beta is positive, it is a increase, when beta is negative it is a decrease.

per <- (y2-y1)/y1 y1*(1+per) y2 > y1*(1+per) [1] 9.810385 > y2 [1] 9.810385

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