Karatzas and Shreve Problem 3.3.38 
Let $X$ be a continuous process and $A$ a continuous, increasing process with $X_0 = A_0 = 0$, a.s.
Suppose that for every $\theta \in \mathbb{R}$, the process
$$Z_t^{\theta} = \exp\left(\theta X_t - \frac{1}{2}\theta^2 A_t\right)$$
is a local martingale. Prove that $X$ is continuous local martingale and quadratic variation of $X$ is $A$.

My attempt :
$$X_t = \ln (Z_t^1)+\frac{1}{2}A_t$$
which means that $X_t$ is continuous semimartingale (since $Z_t^1$ is a local martingale and $A_t$ is of bounded variation). Can some one now help me in proving that $X_t$ is actually a local martingale. Thanks!

First let me say that it was a little more tricky than I previously thought (you need more than only one $\theta$). 
As you have noticed $X_t$ is a continuous semi-martingale, so it has a decomposition into local martingale part $X^m$ plus a finite variation part $X^f$.
Now by Itô on $Z^\theta_t=f_\theta(X_t,A_t)$ you get :
$$\frac{\mathrm dZ^\theta_t}{Z^\theta_t}=\theta\cdot\mathrm X^m_t+\frac{\theta^2}{2}(\mathrm d\langle X\rangle_t-\mathrm dA_t)+\theta\cdot\mathrm dX^f_t.$$
As by hypothesis $Z^\theta_t$ is a continuous local martingale for every $\theta$ then the finite variation part of its sde must be almost surely null. 
So this entails that  $\theta.X^f_t+\frac{\theta^2}{2}(\langle X\rangle_t-A_t)=0$ almost surely. Please notice that $Z^\theta_t>0$ is important.
You get from this first that $X^f_0=0$ so $X^m_0=0$, second if you take $\theta>0 $ and $-\theta$, you get that that $X^f=0$, then third take $\theta=1$ to get $\langle X\rangle_t=A_t$ so that $A_t$ is indeed the quadratic variation of $X$. 
Best regards
