Простой ajax запрос

<script>
      let xhr = new XMLHttpRequest();
      let formData = new FormData();

      formData.append("type", type);
      formData.append("slug", slug);


      xhr.open("POST", '/wp-content/themes/help-consulting/ajax/post-tag.php', true);
    
      xhr.onreadystatechange = function() {
        if (this.readyState != 4) return;
          console.log(this.responseText);
      }

      xhr.send(formData);

</script>