How to interpret regression equations with logarithms, based on log difference being approximate to percentage change? $y = 4 + 2.5\,x + u$
For an increase of 1 unit of $X$ (that is, $X$ to $X+1$), we expect an increase $2.5$ units of $Y$ (that is, $Y$ to $Y+2.5$).
Is that right?

What if there's a/an $\ln$?

$\ln(y) = 1.7 + 0.083\,x + u$
For an increase of 1 unit of $X$ (that is, $X$ to $X+1$), we expect:
a/an $8.3\%$ increase in $Y$ (that is, $Y$ to $Y(1+8.3%)$)?
a/an $0.083\%$ increase in $Y$ (that is, $Y$ to $Y(1+0.083%)$)?
The above is supposed to make use of the fact that log difference is approximately equal to percentage change.

$\ln(y) = 4 + 0.083\,\ln(x) + u$
For a $1\%$ increase in $X$ (that is, $X$ to $X(1+1%)$), we expect:
a/an $8.3\%$ increase in $Y$ (that is, $Y$ to $Y(1+8.3%)$)?
a/an $0.083\%$ increase in $Y$ (that is, $Y$ to $Y(1+0.083%)$)?
The above is supposed to make use of the fact that log difference is approximately equal to percentage change.

Apologies for the confusion everyone. I am wondering about this part in Introductory Econometrics by Wooldridge. It's been a few years since I've interpreted SLR coefficients.
Why is it in the first part $0.083$ becomes $8.3%$ (instead of $0.083%$) but in the second part $0.257$ becomes $0.257%$ (instead of $25.7%$)?




 A: I think you can solve this with exponent and logarithm rules.
For the first model, $y = 4 + 2.5\,x + u$, a $1$-unit increase in $x$ will lead to a $2.5$ increase in $y$, simply because $2.5\,(x + 1) = 2.5\,x + 2.5$.
For the second model, $\ln(y) = 1.7 + 0.083\,x + u$ a $1$-unit increase in $x$ will, by the same token, increase the LHS of the equation $0.083$. But the equation renders the $\ln(y)$, not $y$. Hence, it will be the $\ln(y)$ that will be increased $0.083$. To know the increase in $y$ we will have to exponentiate: $e^{0.083} = 1.086$.
For the third model, $\ln(y) = 4 + 0.083\,\ln(x) + u$, an increase of now (as per the OP) $1\%$ in $x$ will lead to $\ln(1.01\,x)$ as opposed to $\ln(x)$ on the RHS of the equation, or $\ln(1.01\,x)=\ln(1.01)\,+\,\ln(x)$. This will increase the dependent variable by $\ln(1.01)$. But since the dependent variable is in $\log$ scale, we exponentiate to obtain the change in $y$, which is, naturally, $e^{\ln(1.01)}=1.01$.
A: 
For a 1-unit increase in $X$, we expect an 8.3% increase in $Y$

Not quite, because the calculation goes
$$\ln(Y') = 1.7 + 0.083(X + 1) = (1.7 + 0.083X) + 0.083 = \ln(Y) + 0.083$$
so a 1-unit increase in X changes $\ln(Y)$ to
$$\ln(Y) + 0.083 = \ln(Y) + \ln(e^{0.083}) = \ln(e^{0.083}Y) $$
So since  $$e^{0.083} \approx 1.0865$$ it actually leads to about an 8.7% increase in $Y$.
It's worth noting that $e^x \approx 1 + x$ for small values of $x$, so your equation is approximately true, for small values of $0.083$.
I take it you can work out the last example from here.

Does Wooldridge have a typo? The former makes 0.083 into 8.3% and not 0.083% while the latter makes 0.257 into 0.257% and not 25.7%

I don't think so.
If
$$ \ln(Y) = 4.8 + 0.257 \ln(X) $$
then a 1% increase in $X$ gives a new $Y'$ as
$$ \ln(Y') = 4.8 + 0.257 \ln(1.01 X) = \ln(Y) + 0.257 \ln(1.01) $$
We can manipulate the right hand side
$$ \ln(Y) + 0.257 \ln(1.01) = \ln(Y) + \ln(1.01^{0.257}) = \ln(1.01^{0.257} Y) $$
So $Y' = 1.01^{0.257} Y$.  Google says $1.01^{0.257} \approx 1.0026$, which is $0.26\%$, as stated.
A: You can derive all answers yourself by using a simple intuition. Here how it goes. We start with the definition of the derivative:
$$y(x)'\approx\frac{\Delta y(x)}{\Delta x}$$
Next we differentiate both sides of the linear model $y=b_0+b_x x$ and get this:
$$\frac{\Delta y(x)}{\Delta x}\approx b_x $$
$$\Delta y(x)\approx b_x \Delta x$$
Then we say the magic phrase "per unit increase in x", which is the same as saying $\Delta x=1$, and proceed to the simplified formula
$$\Delta y(x)\approx b_x $$
For the log we notice the following:
$$\left(\ln y(x)\right)'=\frac{y'}{y}\approx\frac{\Delta y(x)}{y(x)\Delta x}$$
Then for a model $\ln y=b_0+b_1 x$ we get the following by differentiating both sides of the equation:
$$\frac{\Delta y}{y}\approx b_x \Delta x$$
or for a unit increase in x:
$$\frac{\Delta y}{y}\approx b_x $$
The left hand side can be interpreted as "percent change" but expressed in decimals, so 0.1 would mean 10%.
I'm sure you can pick up from here. It's easy to see what happens when the log is on the right hand side with $\ln x$. It's very similar. Your confusion was in the meaning of the "percent change", which I already explained and shown that it corresponds to terms like $\frac{\Delta s}{s}$
A: You asked a number of questions.  I will paraphrase each, and then answer them.
First question: If $f(x) = 4 + 2.5 x + u$, what is $f(x+1)-f(x)$?  The answer is 2.5, for all $x$.
Second question: If $f(x) = \exp(1.7+0.083 x+u)$, what is $f(x+1)-f(x)$?
$$
\begin{eqnarray}
f(x+1)-f(x) & = & \exp(1.7+0.083(x+1)+u) - \exp(1.7+0.083 x+u) \\
            & = & \exp(1.7+0.083 x+u+0.083) - \exp(1.7+0.083 x+u) \\
            & = & \exp(1.7+0.083 x+u) \cdot \exp(0.083) - \exp(1.7+0.083 x+u) \\
            & = & [\exp(0.083)-1] \cdot \exp(1.7+0.083 x+u) \\
            &\approx&  0.083 \cdot \exp(1.7+0.083 x+u) \\
            &=& 0.083 \cdot f(x) \\
            &=& 8.3\% \cdot f(x)
\end{eqnarray}
$$
Where in the $\approx$ step I used the approximation (which comes from the Taylor series) that for small $x$, $\exp(x) \approx 1+x$.  So if $x$ increases by 1, $f(x)$ increases by 8.3%.  
Third question: Here your question is a little unclear (there's at least one typo), but I think what you want to ask is:  If $f(x) = \exp(4 + 0.083 \ln(x) + u)$, what is $f(1.01 \cdot x)-f(x)$?  This would correspond to the question of how much $f(x)$ changes for a 1% increase in $x$.  
$$
\begin{eqnarray}
f(1.01\cdot x)-f(x) &=& \exp(4 + 0.083 \ln(1.01 x) + u) - \exp(4 + 0.083 \ln(x) + u) \\
               &=& \exp(4 + 0.083 (\ln(1.01)+\ln(x)) + u) - \exp(4 + 0.083 \ln(x) + u) \\
               &=& \exp(4 + 0.083 \ln(x) + u) \cdot \exp(0.083 \ln(1.01)) - \exp(4 + 0.083 \ln(x) + u) \\
               &=& [\exp(0.083 \ln(1.01))-1] \cdot \exp(4 + 0.083 ln(x) + u) \\
               &\approx& [\exp(0.083 \cdot 0.01)- 1] \cdot f(x) \\
               &\approx& (1 + 0.083 \cdot 0.01 - 1) \cdot f(x) \\
               &=& 0.083 \cdot 0.01 \cdot f(x) \\
               &=& 0.083\% \cdot f(x)
\end{eqnarray}
$$
Where at the first $\approx$ I used the approximation that $\ln(1+x) \approx x$ for small $x$, and at the second $\approx$ I again used the approximation that $\exp(x) \approx 1+x$ for small $x$.  (Of course, these are really the same approximation, but one stated in terms of $\exp()$ and one in terms of $\ln()$.)
So the answer is that a 1% increase in $x$ will cause a 0.083% increase in $f(x)$.
Fourth question: Does Wooldridge have a typo?  No.  If you compare the answers to your second and third question, it should now be clear why this is.
