What is the pH of a 0.36 M HCl solution?

can someone help plz? I don't know how to go about figuring this out.

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  1. since it's a strong acid all of the strong acid is converted to [H+] or [H3O+]

    -log[H+]= -log [0.36]= 0,= 0.44

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  2. typo correction (highlighted answer)

    since it's a strong acid all of the strong acid is converted to [H+] or [H3O+]

    -log[H+]= -log [0.36]= 0.44

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  3. Full calculation not give u

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  4. -log(log0.36) how do this Pls full give full calculation

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