The acid catalysed hydrolysis of an compound $\ce{A}$ at $\pu{303K}$: $$\ce {A ->[$k$] B}$$ has a half-life of $\pu{100 min}$ when carried out in a buffer solution of $\mathrm{pH=5}$ and $\pu{10 min}$ when carried out at $\mathrm{pH=4}$. Both the times the half-life are independent of the initial concentration of $\ce{A}$. Find its Rate Law.
I am planning something like this:
Let $$\text{Rate} = k [\ce{A}][\ce{H+}]^x\tag1$$ Now let: $$k'=k[\ce{H+}]^x\tag{constant}$$
And then: $$k' t_\frac 12 = \ln2$$
Is it conceptually correct?
Note:
I have found tons of solutions over the internet all with different answers.
The answer given in my book is x=1 (x is same as is used in eq(1)) and I found the same myself but some sites give answer of the same question as x=2.