The coin is tossed until the first head comes up and the total number of tosses is $X$.
This is a geometric random variable
$X \sim \text{geom}(p)$
With probability mass function
$P(X=x)=(1-p)^{x-1}p$
$H_0: p=0.5$
$H_1: p=0.7$
With equal prior probabilities, the likelihood ratio:
$$ \begin{aligned} \frac{P(H_0|x)}{P(H_1|x)} &= \frac{P(x|H_0)}{P(x|H_1)}\frac{P(H_0)}{P(H_1)} \\\ &= \frac{(1-0.5)^{x-1}(0.5)}{(1-0.7)^{x-1}(0.7)}\frac{0.5}{0.5} \\\ &= \frac{0.5^x}{0.7 \cdot 0.3^{x-1}} \end{aligned} $$
$$\frac{0.5^x}{0.7 \cdot 0.3^{x-1}} = 1 \Rightarrow x \approx 1.7 $$
And the ratio is greater than one for $x > 1.7$
A value of the ratio $> 1$ means that $H_0$ is more strongly supported by the data under consideration than $H_1$.
Hence if $X = 1$ it favors $H_1$, all other outcomes, $X \geq 2$, favours $H_0$.