Given that $a$ is a positive constant and $$\int_{a}^{2a} \frac{t + 1}{t} dt = \ln 7$$ show that $a = \ln k$, where $k$ is a constant to be found. - Edexcel - A-Level Maths Pure - Question 6 - 2017 - Paper 1

Now we need to evaluate it from $a$ to $2a$:

$\left[ t + \ln |t| \right]_{a}^{2a}$

Substituting the limits gives:

$\left( 2a + \ln |2a| \right) - \left( a + \ln |a| \right)$

This simplifies to:

$= 2a - a + \ln |2a| - \ln |a|$

Thus, we have:

$= a + \ln \left( \frac{2a}{a} \right) = a + \ln 2$

We set this equal to $\ln 7$ as given in the problem:

$a + \ln 2 = \ln 7$

To find $a$, we rearrange the equation:

$a = \ln 7 - \ln 2$

Using the properties of logarithms, this becomes:

$a = \ln \left( \frac{7}{2} \right)$

From the previous step, we can identify:

$a = \ln k,$ where $k = \frac{7}{2}$

Thus, we have shown that $a = \ln k$, where $k$ is a constant to be found.